Laxmi

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Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 2 A Shot In the Dark

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11th English Supplementary A Shot In The Dark Warm up
Can you spot 10 differences between these pictures?
How observant are you? Individually, try to spot as many differences as possible in five minutes. If you have spotted less than five, then you really need to improve your observation skill just to save yourself from being misled.

Answer:

Samacheer Kalvi 11th English Solutions Confessions of A Born Spectator Textual Questions

1. Answer the following questions in about 30-50 words each:

A Shot In The Dark Question Answers Question (a)
Why did Philip Sletherby visit Brill Manor?
Answer:
Philip Sletherby visited Brill Manor because he had been invited by Mrs. Honorio Saltpen Jago for a luncheon at Brill Manor.

Shot In The Dark Questions And Answers Question (b)
How would Sletherby gain from his friendship with Honoria Saltpen-Jago?
Answer:
Mr. Philip Sletherby’s hostess was a person of social importance in London, especially of considerable importance in the Chalkshire. He pinned his hopes on her recommendations to get a ticket to contest the MP election in the eastern division of Chalkshire. One of his club acquaintance assured him that Mrs.Saltpen-Jago had a reputation of being an excellent hostess. ‘

A Shot In The Dark Summary In Tamil Question (c)
Why did his travel companion curse and mutter?
Answer:
Mr. Sletherby’s travel companion was a young man of about twenty-two. He cursed and muttered because he had lost his valet along with the family crest. He claimed that he did so foolishly to seal a letter. He had forgotten to collect the valet.

11th English Unit 2 Supplementary Question (d)
Describe Bertie’s problem.
Answer:
Bertie was returning ffomyachting experience for about six months. He had done a damn awkward thing. He left behind his sovereign purse with four quid in it. It was all his worldly wealth at the moment. He must have left the purse on the table. He was stopping at a little country in a near Brondquay for three days’ fishing. As none knows him there, he needed money for his week-end bill and tips and cab to and from the station. In short, he desperately needed three pounds to manage the crisis.

A Shot In The Dark Supplementary Question (e)
‘There was a tinge of coldness in his voice.’ Why?
Answer:
Bertie claimed that he had not seen his mom Mrs. Saltpen-Jago for about six months , because he had gone for yachting. As he had overheard Mr. Sletherby’s intention of visiting Mrs. Saltpen-Jago at Brill Manor he introduced himself as Bertie, Mrs. Saltpen’s younger son. He had lost his sovereign purse. He needed to borrow three quid. As Mr. Sletherby had some suspicion he asked him what was the crest-like? He immediately replied it was demi-lion holding a cross-crosslet. Disbelieving his words, with a tinge of coldness in his voice, Mr. Sletherby said that Mrs. Saltpen-Jago had a grey hound courant on her note paper.

Question (f)
Compare the two crests.
Answer:
The Saltpen crest was “a demi-lion holding a cross-crosslet in its paw.” The Jago crest was a grey hound. Both were used in the family. Bertie could use either of the crests. Bertie had a feeling that he belonged to Saltpens, who used demi-lion crest always.

Question (g)
What was Sletherby’s response to Bertie’s request?
Answer:
Sletherby initially agreed to lend three pounds. But he wanted to ensure that he was not taken for a ride. He asked him if his mother had a close resemblance with him. Bertie claimed that his mom had dark brown hair like himself. Mr. Sletherby was already biased against lending the money as he believed that Grey hound was the crest found on Mrs. Saltpen-Jago’s note book paper. So, he refused to lend the promised loan to Bertie.

Question (h)
What caught his attention about the car door?
Answer:
The car door had two crests along. One was the demi-lion holding a cross-crosslet in its paw and the other was a greyhound.

Question (i)
Describe Honario Saltpen-Jago.
Answer:
Honario Saltpen-Jago was a brilliant and influential woman. She knew exactly when to take up an individual or drop him. She had very fine looks but she spoiled it. She had k beautiful dark brown hair but applied some dye and changed it into blonde.

Question (j)
Why did K.C.’s words shock Sletherby?
Answer:
Mr. Sletherby had left Mr. Bertie, Mrs. Honario Saltpen-Jago’s younger son on his way to Brill Manor in a railway station penniless and frustrated. He had not believed Bertie’s. description of his mom’s dark brown hair and the family’s double crest. When the footman Mr. KC commented on Mrs. Saltpen’s changed Looks, Mr. Sletherby was alarmed and he realized that he had behaved in a mean way. He had mistaken Bertie to be fraud. Thus K.C.’s words had shocked Mr. Sletherby.

2. Rearrange the sentences given below and write a summary of the story in a paragraph. Begin with:

Philip Sletherby was travelling by train to Brill Manor.
(a) The young man introduced himself as Bertie, the son of Saltpen Jago.
(b) Bertie needed three pounds desperately and asked Sletherby to lend it to him.
(c) All the time Sletherby was gazing at the door panel of the car, on which were the two crests – a demi-lion and a grey hound courant.
(d) He was received by Claude People. K.C.,who had kept on talking about various things.
(e) Bertie had left his purse behind, after sealing an envelope with the crest on the purse.
(f) Bertie stated that it was the Jago crest. He further added that his mother’s hair was dark brown similar to his.
(g) While describing the appearance of Honario Saltpen-Jago, K.C. referred to her altered’ hairstyle.
(h) Sletherby realized that Bertie had not lied to him, but that he had mistaken him for a fraud.
(i) He explained that the Saltpen crest was that of a demi-lion.
(j) Sletherby sat dumbstruck on learning that Mrs. Honario Saltpen Jago had changed her dark brown hair to a blonde, just five weeks ago.
(k) Sletherby pointed out to him that his mother’s letter had a greyhound courant crest.
(l) His companion in the train was a yOungman who was searching for something frantically.
(m) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs.Saltpen- Jago was a blonde.

Answers:

(l) His companion in the train was a young man who was searching for something frantically.
(a) The young man introduced himself as Bertie, the son of Saltpen Jago.
(e) Bertie had left his purse behind, after sealing an envelope with the crest on the purse.
(b) Bertie needed three pounds desperately and asked Sletherby to lend it to him.
(i) He explained that the Saltpen crest was that of a demi-lion.
(k) Sletherby pointed out to him that his mother’s letter had a greyhound courant crest.
(J) Bertie stated that it was the Jago crest. He further added that his mother ’s hair was dark brown similar to his. .
(m) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen- Jago was a blonde.
(d) He was received by Claude People. K.C.,who had kept on talking about various things.
(c) All the time Sletherby was gazing at the door panel of the car, on which were the two crests – a demi-lion and a grey hound courant.
(g) While describing the appearance of Honario Saltpen-Jago, K.C. referred to her altered hairstyle.
(j) Sletherby sat dumbstruck on learning that Mrs. Honario Saltpen Jago had changed her dark brown hair to a blonde, just five weeks ago.
(h) Sletherby realized that Bertie had not lied to him, but that he had mistaken him for a fraud.

Answer (for summary of the story in a paragraph)

Philip Sletherby was travelling by train to Brill Manor. His companion was a young man named Bertie, the son of Saltpen Jago. Bertie had left his purse behind after sealing an envelope with the crest on the purse. He asked Sletherby to lend three pounds to him. He also explained that the Saltpen crest was that of a.demi-lion. Sletherby said that his mother’s letter had a greyhound courant crest. Bertie said that his mother’s hair was dark brown similar to his. Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen-Jago was a blonde. He was received by Claude People, K.C., who had kept on talking about various things. While describing the appearance of Honario Saltpen-Jago, K.C. referred to her altered hairstyle and said that she had changed her dark brown hair to a blonde, just five weeks ago. Sletherby realized that Bertie had not lied to him, but that he had mistaken him for a fraud.

Additional Questions

1. Rearrange the sentences given below:

(a) Unable to arrest Bob himself, he sends a plainclothed man to do it.
(b) Bob kept his words and reached the spot before 10 p.m. and waited anxiously.
(c) A patrolman (Jimmy wells) finds a notorious criminal wanted by Chicago police is Bob.
(d) Both promised to meet at the same spot at 10 pm 20 years later.
(e) Jimmy wells and Bob had dinner at Big Joe’ Brady’s restaurant.

Answer:

(e) Jimmy wells and Bob had dinner at Big Joe’ Brady’s restaurant.
(d) Both promised to meet at the same spot at 10 pm 20 years later.
(b) Bob kept his words and reached the spot before 10 p.m. and waited anxiously.
(c) A patrolman (Jimmy wells) finds a notorious criminal wanted by Chicago police is Bob.
(a) Unable to arrest Bob himself, he sends a plainclothes man to do it.

2. Rearrange the sentences given below:

(a) While describing the appearance of Mrs. Saltpen-Jago, K.C. described her altered hairstyle but Sletherby was in a daze after seeing the double crest on the door of the car.
(b) Sletherby realized that Bertie had not lied, but that he had mistaken him for a fraud.
(c) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen- Jago was a blonde.
(d) When Sletherby referred to a greyhound courant, Bertie stated that it was Jago crest. He further added that his mom had dark brown hair similar to his.
(e) Bertie having lost his purse wanted a loan of three pound, from Mr. Sletherby.

Answers:

(e) Bertie having lost his purse wanted a loan of three pound, from Mr. Sletherby.
(d) When Sletherby referred to a greyhound courant, Bertie stated that it was Jago crest. He further added that his mom had dark brown hair similar to his.
(c) Suspecting foul play, Sletherby did not give him any money, as he knew Mrs. Saltpen-Jago was a blonde.
(a) While describing the appearance of Mrs. Saltpen-Jago, K.C. described her altered hairstyle but Sletherby was in a daze after seeing the double crest on the door of the car.
(b) Sletherby realized that Bertie had not lied, but that he had mistaken him for a fraud..

3 Explain the following phrases selected from the story in your own words and work with a partner to make sentences using these phrases:

Question (a)
on approval – on acceptance
Answer:
Seetha went abroad on her dad’s approval.

Question (b)
a rustic holiday – a vacation to be spent in a village
Answer:
The students who went on a rustic holiday to Kurangani forest were caught in a forest fire accident.

Question (c)
a double-distilled idiot – very stupid person
Answer:
Some double – distilled idiots ill-treat their parents in old age.

Question (d)
a tinge of coldness – a shade of hostility lack of friendliness
Answer:
The manager turned down the worker’s plea for a pay hike with a tinge of coldness.

Question (e)
making up for lost time – compensate
Answer:
The boy had to study two hours more everyday to make up for the lost time.

(4) Answer the following questions in a paragraph of about 100-150 words each.

Question (a)
Describe the youth’s strange behaviour when he was in the train.
Answer:
The young man who entered the coach gave out a smothered curse. He was engaged in searching something elusive angrily and uselessly. From time to time, he dug a six penny bit out of a waist coat pocket and stared at it sadly, then resumed his search. He voluntarily broke the silence. He exclaimed that Mr. Sletherby was-going to Bill Manor. He introduced himself as Bertie, the younger son of Mrs. Saltpen-Jago. He admitted that he was away for about six months and had not seen his own mother. Making use of the lucky coincidence that he was going to Brill Manor, he asked for a loan of three pounds as he had lost his sovereign purse and was desperately in need of help. He promised to meet him on the subsequent Monday.
“There is no need to suffer silently and there is no shame in seeking help.”

Question (b)
One has to be cautious and not be carried away by stories. How did Sletherby exhibit caution?
Answer:
Sletherby is a very cautious person. From the beginning he had decided not to be carried away by the ‘stories’ of the young man. He laid the trap for the young man by feigning absolute ignorance about his mom’s looks. He asked him if his mom resembled him. Bertie replied that his mom had dark brown hair like himself. Besides, in his mental eye, Sletherby compared the crest he found in the invitation received from Mrs. Saltpen-Jago “a grey hound” with the young man’s description of the royal crest “demi-lion holding a cross-crosslet in its paw”. The two details contradicted Mr. Sletherby’s knowledge of Mrs. Saltpen-Jago’s royal family. There is something basically inhuman not to be interested in the affairs of a fellow human being “in distress”. .
“The truth is a beautiful and terrible thing and hence should be treated with caution.”

Question (c)
How did Sletherby’s judgement of Bertie turn out to be a wrong one?
Answer:
Sletherby concluded that Bertie was a fraud. He compared his looks with that of his mom. The footman of Mrs.Saltpen-Jago while describing her appearance, said that she had altered her looks only five weeks ago. She had changed her dark brown hair to a blonde. Sletherby also saw the double crest of the royal family on the door panel of the car which displayed a demi-lion and a greyhound courant. The evidences which made Mr. Sletherby believe that Bertie must have been a fraud turned out to be solid evidences ‘ indicating Bertie’s genuineness as a legal heir of Mrs. Saltpen-Jago. Sletherby realized

with alarm that Bertie had not lied to him, but he had mistaken an honest man in trouble to be an impostor. Sletherby’s political future was heading to a disaster simply because he refused to lend three pounds to a young man in trouble. He was penny wise and pound foolish.

“Before you assume, learn the facts. Before you judge, understand why. ”

Question (d)
As Sletherby, would you apologise to Bertie for your rude behaviour? Give reasons.-
Answer:
Of course, yes. I would earnestly apologise for mistaking Bertie to be a fraud. I would explain the circumstantial evidences which really worked against him. The crest of a grey hound I had found in his mom’s letter and the present altered colour of his mom’s hair had clouded my judgement against him. I would admit that I could have offered at least two pounds to help him overcome the problem caused by his carelessness. I would express my regrets for leaving him behind furious and disappointed.
“Life becomes easier when you learn to accept the apology you never got. ”

Question (e)
‘Seeing is believing’. How is this humorously disproved in this story? Bring out the irony in the situation. .
Answer:
“Seeing is believing” is an age old adage which is also contradicted by Plato’s saying Appearance is always deceptive because it appears to appear”. Some times truth is not quite impressive. She needs the support of limping old maid called time to prove herself. Being a cynic and skeptic, Mr. Sletherby takes up a stand right at the beginning of the encounter, not to show any interest in the young man’s fussy outbursts. He carefully lays the trap by telling a lie that he had not seen his mom. He further asks him cleverly if his mom resembled him.

Both Bertie himself and Mr. Sletherby are unaware of the altered looks of Mrs. Saltpen-Jago. There is a dramatic irony when the son himself gives a different version of his mom’s appearance. This influences the decision of Mr. Sletherby in refusing to lend Bertie a loan of three pounds. Thus the adage “Seeing is believing” is humorously disproved in this story. Enquiring thoroughly and compassionately one can find the truth.
“Everything we hear is an opinion, not a fact.
Everything we see is a perspective, not the truth.

Additional Questions

Question (a)
Attempt a character sketch of Mr. Philip Sletherby.
Answer:
Mr. Philip Sletherby was a politician but much known to the people of Chalkshire. He wishes to contest for a MP seat in Chalkshire with the recommendation of a level-headed woman namely Mrs. Saltpen-Jago. Fortunately he had been invited to spend the weekend with her at Brill Manor. He believes that his political pilgrimage would turn out to be a blessing. There is something mean about his unwillingness to show any interest in the affairs of an angry young travelling companion. Mr. Philip Sletherby loves things and uses people to advance his career prospects. At the very beginning, he doubts the credentials of Bertie who introduces himself as the younger son of Bertie. He is quick to judge people. Though the young man wanted only three pounds, he quizzes him with many questions about his royal crest and contradicts him with his ice-cold logic. He is proud of having laid a trap saying he had never seen his mother.

He offers to assist the young man in distress. But when Bertie wants to leave, he refuses to part with money saying that his mom was a pronounced blonde. He prided himself for his cleverness. But he is shocked to see the double crest on the car’s door panel, demi-lion holding a cross-crosslet in his paw and a greyhound. He is petrified to leam from K.C.’ that Mrs. Saltpen-Jago had recently dyed her beautiful dark brown hair blonde. Mr. Sletherby realized that his cleverness has, in reality, turned out to be his meanness. He becomes troubled that his political future may be lost due to his three pound de-buckle. He has proved himself penny wise and pound foolish.
“You must make the choice, to take the chance if you want anything, in life to change. ”

Question (b)
Attempt a character sketch of Mrs. Saltpen-Jago.
Answer:
Mrs. Saltpen is a brilliant woman. She is level headed. She is a clear thinker. She knows exactly when to take up an individual or a social cause. Being a lady of discernment, she may choose to drop a person if she finds him unworthy of her attention. She is an influential woman. But occasionally, she is eccentric. She spoils herself and her chance by being too restless. No wonder Bertie too is restless like his mom. She is never calm and composed. She used to have beautiful dark brown hair which jelled well with her fresh complexion.

Recently she has electrified people by dying it blonde. It has in-fact ruined her looks. K.C. unravels many details about Mrs. Saltpen-Jago. It is obvious that her influence in Chalkshire is more due to her affluence and large estate than her personality traits. Her inconstant nature is obvious in her ability to raise someone from the gutters to the top or drop someone she does not like. The only thing charming about her is that she treats her guests well as a good hostess.

I. Choose the right option.

Question 1.
Philip Sletherby was on a profitable pilgrimage to ______
(a) Mecca
(b) Israel
(c) Velankanni
(d) Brill Manor
Answer:
(d) Brill Manor

Question 2.
Mrs ______ was a person of some social importance in London.
(a) Thatcher
(b) Victoria
(c) Williams
(d) Saltpen-Jago
Answer:
(d) Saltpen-Jago

Question 3.
Sletherby hostess had considerable influence in the country of ______
(a) Cannibals
(b) Yorkshire
(c) Chalkshire
(d) Scotland
Answer:
(c) Chalkshire

Question 4.
______ was the second son of Mrs. Saltpen-Jago.
(a) Bob
(b) Jimmy Wells
(c) Bertie
(d) James
Answer:
(c) Bertie

Question 5.
Mr. ______ was invited to spend the weekend with Mrs. Saltpen-Jago.
(a) Bertie
(b) Claude
(c) KC
(d) Philip Sletherby
Answer:
(d) Philip Sletherby

Question 6.
Bertie claimed that he had not seen his mother for a period of ______ months.
(a) three
(b) six
(c) four
(d) two
Answer:
(b) six

Question 7.
Bertie was ______ when his mom was in town.
(a) fishing
(b) yachting
(c) travelling
(d) skiing
Answer:
(b) yachting

Question 8.
The young man confessed that he had done an ______ thing.
(a) ugly
(b) awkward
(c) embarrassing
(d) efficient
Answer:
(b) awkward

Question 9.
Bertie claims to have lost his sovereign purse with ______ quid in it.
(a) ten
(b) six
(c) four
(d) five
Answer:
(c) four

Question 10.
Bertie compared himself to a double distilled ______ for leaving his purse behind.
(a) genius
(b) fool
(c) idiot
(d) clown
Answer:
(c) idiot

Question 11.
Bertie planned to spend three days at ______
(a) Brazil
(b) Brondquay
(c) Mumbai
(d) Ceylon
Answer:
(b) Brondquay

Question 12.
Bertie wanted to go for ______ for three days.
(a) trekking
(b) fishing
(c) reading
(d) bungee jumping
Answer:
(b) fishing

Question 13.
Initially Sletherby gave Bertie hopes by agreeing to lend him ______ pounds.
(a) one
(b) two
(c) three
(d) four
Answer:
(c) three

Question 14.
As suspicion crept into Sletherby’s mind about the genuineness of the boy’s claim there was a tinge of ______ in his voice.
(a) pity
(b) anger
(c) coldness
(d) love
Answer:
(c) coldness

Question 15.
Bertie said that his mom had dark ______ hair.
(a) red
(b) brown
(c) black
(d) blond
Answer:
(b) brown

Question 16.
Slether by found Bertie’s mother having ______ hair.
(a) brown
(b) red
(c) black
(d) blond
Answer:
(d) blond

Question 17.
Sletherby congratulated himself on his ______
(a) meanness
(b) judgement
(c) astuteness
(d) cautiousness
Answer:
(c) astuteness

Question 18
______ wanted to play a round of golf with Sletherby.
(a) Mrs. Saltpen-Jago
(b) Peter
(c) Claude People
(d) Bertie
Answer:
(c) Claude People

Question 19.
Sletherby’s eyes were fixed on the ______ on which were displayed two crests: a greyhound courant and a demi-lion holding in its paw a cross-crosslet.
(a) tyre
(b) mirror
(c) door panel
(d) bonnet
Answer:
(c) door panel

Question 20.
The idiotic change Mrs. Saltpen-Jago had made five weeks ago was ______
(a) plastic surgery of her nose
(b) dying her dark brown hair blonde
(c) dying her brown hair white
(d) cutting her long curly hair like a school girl
Answer:
(b) dying her dark brown hair blonde

II. Identify the speaker.

Question 1.
“Oh, you’re staying with Mrs. Saltpen-Jago for the week-end.
Answer:
Club acquaintance of Mrs. Sletherby.

Question 2.
“Didn’t I hear you say you were going down to stay with Mrs. Saltpen-Jago at Brill Manor?
Answer:
Bertie to Philip Sletherby.

Question 3.
“I’ve done a damned awkward thing.”
Answer:
Bertie

Question 4.
“What’s your crest by the way?”
Answer:
Sletherby to Bertie.

Question 5.
“I think I can manage that.”
Answer:
Philip Sletherby.

Question 6.
“Thanks awfully. It’s jolly good of you.”
Answer:
Bertie to Sletherby

Question 7.
“The demi-lion is the Saltpen crest.”
Answer:
Bertie to Sletherby

Question 8.
“My station is the next one.”
Answer:
Bertie to Sletherby

Question 9.
“Does she resemble you at all in feature?”
Answer:
Sletherby to Sletherby

Question 10.
“She has the same dark brown hair.”
Answer:
Bertie about his mom to Sletherby

Question 11.
“You’ve forgotten the three quid.”
Answer:
Bertie to Sletherby

Question 12.
“I’ve no intention of lending you three pounds, or three shillings.”
Answer:
Sletherby to Bertie

Question 13.
“She is a pronounced blonde.”
Answer:
Sletherby about Mrs. Saltpen-Jago to Bertie

Question 14.
“Hullo, Sletherby! You spending the week-end at Brill?”
Answer:
Claude People K.C.

Question 15.
“We’ll have a round of golf together.”
Answer:
Claude People K.C,

Question 16.
“Good appearance, until she made that idiotic change.”
Answer:
Claude People K.C.

Question 17.
“Change?…. What change?”
Answer:
Sletherby to K.C.

Question 18.
“She used to have beautiful dark brown hair, which went very well with her fresh complexion” ,
Answer:
Claude People K.C. to Sletherby

Question 19.
“Five weeks ago, she electrified everybody by appearing as a brilliant blonde.”
Answer:
Claude People K.C. to Sletherby

Question 20.
“I say, what’s the matter with you? You look rather ill.”
Answer:
Claude People K.C. to Sletherby.

III. Reading Comprehension.

Question 1.
He had scarcely glanced at a couple of pages, however,, when a smothered curse caused him to glance hastily at the only other occupant of the carriage. His travelling companion was a young man of about two-and-twenty, with dark hair, fresh complexion, and the blend of smartness and disarray that marks the costume of a ‘nut’ who is bound on a rustic holiday. He was engaged in searching furiously and ineffectually for some elusive or non-existent object; from time to time he dug a sixpenny bit out of a waistcoat pocket and stared at it ruefully, then recommenced the futile searching operations. A cigarette-case, matchbox, latchkey, silver pencil case, and railway ticket were turned out on to the seat beside him, but none of these articles seemed to afford him satisfaction; he cursed again, rather louder than before. The vigorous pantomime did not draw forth any remark from Sletherby, who resumed his scrutiny of the magazine.

Question (a)
What made Sletherby look at his fellow traveller?
Answer:
The young travelling companion let out a smothered curse. This made him look at the young travelling companion.

Question (b)
What, did Mr. Sletherby deduce from the looks of the young man?
Answer:
The young man must be around twenty two. He was bound for a rustic holiday.

Question (c)
What was the young man doing inefficiently?
Answer:
The young man was furiously searching for some elusive or non-existent object.

Question (d)
What did the young man’s rueful stare at the six penny imply?
Answer:
The young man’s rueful stare at the six penny implied the young man was in a financial . trouble.

Question (e)
Did the vigorous pantomime get the young man any help? Why?
Answer:
The vigorous pantomime turned out to be useless. Mr. Sletherby did not evince any interest in knowing the young man’s trouble.

2. “Not exactly, but left behind, which is almost as bad; just as inconvenient, anyway. I’ve come away .without my sovereign-purse, with four quid in it, all my worldly wealth for the moment. It was in my pocket all right, just before I was starting, and then I wanted to seal a letter, and
the sovereign-purse happens to have my crest on it, so I whipped it out to stamp the seal with, and, like a double-distilled idiot, I must have left it on the table. I had some silver loose in my pocket, but after I’d paid for a taxi and my ticket I’d only got this forlorn little six pence left. I’m stopping at a little country inn near Brondquay for three days’ fishing; not a soul knows me there, and my week-end bill, and tips, and cab to and from the station, and my ticket on to

Brill, that will mount up to two or three quid, won’t it? If you wouldn’t mind lending me two pound ten, or three for preference, I shall be awfully obliged. It will pull me out of no end of a hole.”

Question (a)
What was left behind?
Answer:
The young man’s sovereign purse was left behind.

Question (b)
What according to the young man, was in the purse?
Answer:
There was four quid in the purse.

Question (c)
What was the need to open his purse?
Answer:
He had to take out his crest to seal a letter.

Question (d)
Why did the young man call himself a distilled idiot?
Answer:
The young man had left his purse on the table while sealing the letter. So, he called himself a double distilled idiot.

Question (e)
“Haste makes waste.” Relate this saying to the young man’s loss.
Answer:
The young man was probably in a hurry to hire a cab and reach the railway station. In a haste he forgot to collect the purse he had kept on the table. Thus, it is true that haste makes waste.

3. “Thanks awfully. It’s jolly good of you. What a lucky thing for me that I should have chanced across one of the mater’s friends. It will be a lesson to me not to leave my exchequer lying about anywhere, when it ought to be in my pocket. I suppose the moral of the whole thing is don’t try and convert things to purposes for which they weren’t intended. Still, when a sovereign-purse has your crest on it-”
“What is your crest, by the way?” Sletherby asked, carelessly.

“Not a very common one,” said the youth; “a demi-lion holding a cross-crosslet in its paw.” “When your mother wrote to me, giving me a list of trains, she had, if I remember rightly, a greyhound *courant on her notepaper,” observed Sletherby. There was a tinge of coldness in his voice.

“That is the Jago crest,” responded the youth promptly; “the demi-lion is the Saltpen crest. We have the right to use both, but I always use the demi-lion, because, after all, we are really Saltpens.”

Question (a)
Why did Bertie thank Mr. Sletherby?
Answer:
He thanked Bertie hoping that he would extend him a loan of three pounds.

Question (b)
What was uncommon about the crest Bertie described?
Answer:
It was a demi-lion holding a cross-crosslet in his paw. Thus, it was an uncommon crest.

Question (c)
How did Mr. Sletherby’s observation drive Bertie to defend the genuineness of his crest?
Answer:
Mr. Sletherby observed that in the note paper he had received from Mrs. Saltpen-Jago, he had seen “Greyhound courant” crest printed. This drove Bertie to defend his crest saying his mom used Jago crest whereas he used Saltpen’s crest.

Question (d)
How did Bertie happen to use a different crest?
Answer:
Bertie used Saltpen’s crest. His mom used Jago crest i.e., Greyhound. Children had the option to use either Saltpen’s or Jago crests.

4. “Good-bye,” said Sletherby.

“You’ve forgotten the three quid,” said the young man, opening the carriage-door and pitching his suit-case on to the platform.

“I’ve no intention of lending you three pounds, or three shillings,” said Sletherby severely. “But you said-”
“I know I did. My suspicions hadn’t been roused then, though I hadn’t necessarily swallowed your story. The discrepancy about the crests put me on my guard, not withstanding the really brilliant way in which you accounted for it. Then I laid a trap for you; I told you that I had never met Mrs. Saltpen-Jago. As a matter of fact I met her at lunch on Monday last. She is a pronounced blonde.”

The train moved on, leaving the soi-disant cadet of the Saltpen-Jago family cursing furiously on the platform.

Question (a)
Did Mr, Sletherby really forget three quids. How?
Answer:
No, Sletherby did not forget three quid loan demanded by the young man. He simply took him to be a fraud and refused to lend him even three penny.

Question (b)
What had put Mr. Sletherby on guard?
Answer:
The information the boy disclosed about the family crest of Saltpen-Jago put Mr. Sletherby on guard.

Question (c)
What lie did Mr. Sletherby use to trap the young man?
Answer:
He had never seen Bertie’s mom Mrs. Saltpen-Jago.

Question (d)
Which claim of Bertie about his mom’s looks worked against him?
Answer:
The claim that his mom had dark brown hair similar to him worked against him. This . statement made Mr. Sletherby conclude that Bertie was a fraud.

Question (e)
What did Bertie do to express his feelings on the disappointment caused by Mr. Sletherby?
Answer:
Bertie cursed Sletherby furiously stamping his feet on the platform.

5. Sletherby heard not a single word, noted not one of the details that were being expounded to him. His eyes were fixed on the door panel, on which were displayed two crests: a greyhound courant and a demi-lion holding in its paw a cross-crosslet. “Brilliant woman, level-headed, a clear thinker, knows exactly when to take up an individual or a cause, exactly when to let him or it drop. Influential woman, but spoils herself and her chances by being too restless. No repose. Good appearance, too, till she made that idiotic change.”

“Change?” queried Sletherby, “what change?”

“What change? You don’t mean to say- Oh, of course, you’ve only known her just lately. She used to have beautiful dark brown hair, which went very well with her fresh complexion; then one day, about five weeks ago, she electrified everybody by appearing as a brilliant blonde. Quite ruined her looks. Here we are. I say, what’s the matter with you? You look rather ill.”

Question (a)
Why was Mr. Sletherby speechless?
Answer:
Mr. Sletherby saw two crests on the car’s door panel. One was a grey bound and the other was what Bertie had described, a demi-lion holding in its paw a cross, crosslet. This made him speechless.

Question (b)
Who described Mrs. Saltpen-Jago as a brilliant and influential woman?
Answer:
Mr. K. C. described Mrs. Saltpen-Jago as a brilliant and influential woman.

Question (c)
How did Mrs. Saltpen-Jago spoil her chances?
Answer:
She spoiled her chances by being too restless.

Question (d)
What idiotic change had Mrs. Saltpen-Jago made in her good appearance?
Answer:
She dyed her beautiful dark brown hair into blonde. This idiotic change ruined her looks.

Question (e)
Why did Mr. Sletherby look very ill suddenly?
Answer:
Mr. Sletherby had doubted Bertie’s crest and his claim of brown hair of his mom. But he realized with alarm that Bertie had told the truth about the crest and about his mom’s hair. He looked ill as he had been too mean to Mrs. Salt-pen’s son.

A Shot In the Dark About the Author

Hector Hugh Munro (1870-1916) better known by his pen name Saki was a brilliant British writer. His witty, mischievous and sometimes macabre stories, satirize Edwardian society and culture. He is considered a master of short story and often compared to Q, Henry. His stories also tend to end with a twist like those of O. Henry’s.

A Shot In the Dark Summary

Philip Sletherby travels in a train to meet Mrs. Saltpen Jago. He decided to gain some political mileage by getting her acquaintance. Sletherby wanted to be a candidate for the MP election in the eastern division of Chalkshire. As she had invited him on a friendly luncheon-party at her country house he counted on her for nominating him. The silence in the compartment was spoiled by the smothered cries of a young man who was obviously looking for something. As Mr, Philip Sletherby did not show any interest in his despair, he himself explained his situation. He was glad that the narrator was going to spend his week-end with his mother Mrs. Saltpen – Jago at Brill Manor. The boy said that he hadn’t seen his mom for almost six months. He was Bertie and he had lost his royal seal and the money purse. He needed three quid for his brief stay and ticket fare to his home. He was on a fishing expedition.

Mr. Sletherby agreed to lend but asked him, what was his crest? The young man said it was the demi-lion holding a eross-crosslet in its paw, With a tinge of coldness in his voice, Sletherby said that his mom had written a grey hound courant on her notepaper. The young man said they could use both. Demi-lion was Saltpen crest and Grey Hound was Jago-crest. The young man said that he always used the demi-lion. He also added that he was getting down in the next station. Sletherby asked him if his mom resembled him. The boy said that she had the same dark brown hair like him. When the boy was about to get out, he reminded him of the loan of three quid. Sletherby said that he had met his mom who was a pronounced blonde. The young man stamps his. feet furiously. A tall footman had arrived at the railway station to receive him.

A luxury vehicle was parked. The footman Mr. K.C. explained a lot about the car. But nothing entered his brain as he saw two crests on the car “s door panel-a grey hound courant and a demi-lion holding in its paw a cross-erosslet. The second blow follows when KC said that Mrs. Saltpen-Jago used to have beautiful dark brown hair which agreed well with her complexion. She shocked everybody bv changing her looks as a brilliant blonde. She had ruined her looks. Suddenly Mr. Sletherby turned pale. Mr. K.C. asked if he was ill.

1. Rearrange the sentences given below:

astuteness – cleverness
blonde – pale yellow hair
chaise – horse-drawn carriage
*courant -/ko ra:nt/ – animal represented in the act Of running
crest -a symbol representing a family, reproduced on writing paper
discrepancy – difference
embarked – boarded
espied – caught sight of
exchequer – national treasury
expounded – explained
Honoria /hD ‘ na:na/ – (Latin) honour, a woman of reputation
ineffectually – In vam
mater / ‘ melta/ – (Latin) mother
pantomime – exaggerated behaviour
quid – (British) one pound Sterling
ruefully – regretfully, sorrowfully
sedately – in a dignified manner
soi- disant -self styled
sumptuous – expensive looking
tackle – equipment required for a task or sport
volubility – fluency, talkativeness

Additional:
apologise – seek forgiveness
alarmed – afraid of danger
dumb struck – shocked
exactly – correctly, accurately
frantically – quickly in a disorderly manner

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Supplementary Chapter 2 A Shot In the Dark Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Students can Download Computer Science Chapter 7 Composition and Decomposition Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 7 Composition and Decomposition

Samacheer Kalvi 11th Computer Science Composition and Decomposition Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 7 Book Back Answers Question 1.
Suppose u, v = 10, 5 before the assignment. What are the values of u and v after the sequence of assignments?
1. u : = v
2. v : = u
(a) u, v = 5, 5
(b) u, v = 10, 5
(c) u, v = 5, 10
(d) u, v = 10, 10
Answer:
(a) u, v = 5, 5

Composition And Decomposition In Computer Science Question 2.
Which of the following properties is true after the assignment (at line 3?
1. – – i + j = 0
2. i, j : = i + 1, j – 1
3. – – ?
(a) i + j > 0
(b) i + j < 0
(c) i + j = 0
(d) i = j
Answer:
(c) i + j = 0

Class 11 Computer Science Chapter 7 Solutions Question 3.
If C1 is false and C2 is true, the compound statement

(a) S1
(b) S2
(c) S3
(d) none
Answer:
(b) S2

Computer Science Samacheer Kalvi Question 4.
If C is false just before the loop, the control flows through

(a) S1 ; S3
(b) S1 ; S2 ; S3
(c) S1 ; S2 ; S2 ; S2 ; S3
(d) S1 ; S2 ; S2 ; S2 ; S3
Answer:
(a) S1 ; S3

Samacheer Kalvi Computer Science Question 5.
If C is true, S1 is executed in both the flowcharts, but S2 is executed in

(a) (1) only
(b) (2) only
(c) both (1) and (2)
(d) neither (1) nor (2)
Answer:
(a) (1) only

Samacheer Kalvi Guru 7th Computer Science Question 6.
How many times the loop is iterated?
i : = 0
while i ≠ 5
i : = i + 1
(a) 4
(b) 5
(c) 6
(d) 0
Answer:
(b) 5

PART – 2
II. Short Answers

Samacheer Kalvi Guru 11th Computer Science Question 1.
Distinguish between a condition and a statement.
Answer:

Samacheer Kalvi Computer Science Book Question 2.
Draw a flowchart for conditional statement.
Answer:

11 Computer Science Samacheer Kalvi Question 3.
Both conditional statement and iterative statement have a condition and a statement. How do they differ?
Answer:
Conditional Statement:

• Statements are executed only once when the condition is true.
• If condition statement.

Iterative Statement:

• Iterative statement repeatedly evaluates a condition and executes a statement until it becomes false.
• While condition statement.

Samacheer Kalvi Guru Computer Science Question 4.
What is the difference between an algorithm and a program?
Answer:
Algorithm:

• Algorithm is for human readers to understand.
• Knowledge of English is needed.
• Easy to understand.

Program:

• Program is for the computers to execute directly.
• Knowledge of programming language is required.
• It is difficult to understand.

Computer Science Textbook Solutions Question 5.
Why is function an abstraction?
Answer:
Once a function is defined, it can be used over and over and over again. Reusability of a single function several times is known as an abstraction.

Computer Chart For 11th Class Question 6.
How do we refine a statement?
Answer:
In refinement, each statement is repeatedly expanded into more detailed statements in the subsequent levels.

PART – 3
Explain in Brief

Computer Charts For Class 11 Question 1.
For the given two flowcharts write the pseudo code.
Answer:

1. Enter A,B
2. Initialise Q = 0, r = A
3. If r > B, then do Q = Q + 1; r = r – B else r, q
4. Exit

Samacheer Kalvi Guru 11 Computer Science Question 2.
If C is false in line 2, trace the control flow in this algorithm.

1. S1
2. – – C is false
3. if C
4. S2
5. else
6. S3
7. S4

Answer:
S1 ; S2 ; S4

Statement Meaning In Computer Question 3.
What is case analysis?
Answer:
Case Analysis statement generalizes the problem into multiple cases. Case Analysis splits the problem into an exhaustive set of disjoint cases.

Question 4.
Draw a flowchart for -3 case analysis using alternative statements.
Answer:

Question 5.
Define a function to double a number in two different ways:

1. n + n
2. 2 x n

Answer:
1. Double (n)
– – inputs: n is a real number or an integer, n > 0
– – Outputs: y is a real number or an integer such that y = n + n

2. Double (n)
– – inputs: n is a real number or an integer, n > 0
– – Outputs: y is a real number or an integer such that y = 2 x n

PART – 4
IV. Explain in Detail

Question 1.
Exchange the contents: Given two glasses marked A and B. Glass A is full of apple drink and glass B is full of grape drink. Write the specification for exchanging the contents of glasses A and B, and write a sequence of assignments to satisfy the specification.
Answer:
Let the variables a, b, c represent Glass A, Glass B and Glass C and a, b, c can store values APPLE, GRAPE or EMPTY.
Specification:

1. exchange (a, b)
2. – – inputs: a, b : = APPLE, GRAPE
3. – – outputs: a, b : = GRAPE, APPLE

Algorithm:

Question 2.
Circulate the contents: Write the specification and construct an algorithm to circulate the contents of the variables A, B and C as shown below: The arrows indicate that B gets the value of A, C gets the value of B and A gets the value of C.

Specifications:

1. circulate
2. – – inputs: a, b, c: = A, B, C
3. – – outputs: a, b, c: = C, A, B

Algorithm:

1. circulate (a, b, c)
2. – – a, b, c: = A, B, C
3. temp : = c
4. c : = b
5. b : = a
6. a : = temp
7. – – a, b, c: = C, A, B

Question 3.
Decanting problem. You are given three bottles of capacities 5, 8, and 3 litres. The 8L bottle is filled with oil, while the other two are empty. Divide the oil in 8L bottle into two equal quantities. Represent the state of the process by appropriate variables. What are the initial and final states of the process? Model the decanting of oil from one bottle to another by assignment. Write a sequence of assignments to achieve the final state.
Answer:
(a)
Let a, b, c be the variables whose maximum value are 8L, 5L and 3L respectively.
Initial State:

a, b, c: = 8, 0, 0

Final State:

a, b, c: = 4, 4, 0

Specifications:

1. decant
2. – – inputs: a, b, c : = 8, 0, 0
3. – – outputs: a, b, c : = 4, 4, 0

Algorithm:
Let us assume that a: = b denote oil in b is poured into a bottle until either “a” is full or “b” becomes empty.

1. decant (a, b, c)
2. – – a, b, c : = 8, 0, 0
3. b : = a
4. – – a, b, c : = 3, 5, 0
5. c : = b
6. – – a, b, c : = 3, 2, 3
7. a : = c
8. – – a, b, c : = 6, 2, 0
9. c : = b
10. – – a, b, c : = 6, 0, 2
11. b : = a
12. – – a, b, c : = 1, 5, 2
13. c : = b
14. – – a, b, c : = 1, 4, 3
15. a : = c
16. – – a, b, c: = 4, 4, 0

Question 4.
Trace the step – by – step execution of the algorithm for factorial(4).
factorial(n)
– – inputs : n is an integer , n ≥ 0
– – outputs : f = n!
f, i : = 1,1
while i ≤ n
f, i : = f x i, i + 1
Answer:

Samacheer kalvi 11th Computer Science Composition and Decomposition Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
Which one of the following is odd?
(a) Python
(b) C++
(c) C
(d) Ctrl + S
Answer:
(d) Ctrl + S

Question 2.
………………. is a diagrammatic notation for representing algorithms.
(a) Pseudo code
(b) Flowchart
(c) Program
(d) Languages
Answer:
(b) Flowchart

Question 3.
There are important control flow statements.
(a) four
(b) three
(c) two
(d) five
Answer:
(b) three

Question 4.
A ………………. statement is composed of a sequence of statements.
(a) iterative
(b) conditional
(c) sequential
(d) alternative
Answer:
(c) sequential

Question 5.
A ………………. is contained in a rectangular box with a single outgoing arrow, which points to the box to be executed next.
(a) statement
(b) composition
(c) notation
(d) condition
Answer:
(a) statement

Question 6.
The triangle is right – angled, if ……………….
(a) C = a – b
(b) C² = a² + b²
(c) C² = (a + b)²
(d) c² = a² – b²
Answer:
(b) C² = a² + b²

Question 7.
The algorithm can be specified as ……………….
(a) monochromatize (a, b, c)
(b) a = b = 0
(c) C = A + B + C
(d) none
Answer:
(a) monochromatize (a, b, c)

Question 8.
After an algorithmic problem is decomposed into subproblems, we can abstract the subproblems as ……………….
(a) refinement
(b) pseudo – code
(c) decomposition
(d) functions
Answer:
(d) functions

Question 9.
Which one of the following is the elementary problem solving techniques?
(a) Specification
(b) Abstraction
(c) Composition
(d) decomposition
Answer:
(d) decomposition

Question 10.
Which one of the following have only high level details?
(a) Flow chart
(b) Algorithm
(c) Programs
(d) Pseudocode
Answer:
(b) Algorithm

Question 11.
How many different notations are there for representing algorithms?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 12.
Which one of the following notations will be executed by computers?
(a) Flow chart
(b) Pseudocode
(c) Programming languages
(d) Compiler
Answer:
(c) Programming languages

Question 13.
Which one of the following algorithmic notation is used for communication among people?
(a) Flow chart
(b) Pseudo code
(c) PL
(d) Interpreter
Answer:
(b) Pseudo code

Question 14.
Which one of the following algorithmic notation is used for giving visual intuition of control flow?
(a) Flow chart
(b) Programming languages
(c) Pseudo code
(d) Compiler
Answer:
(a) Flow chart

Question 15.
The algorithmic notation similar to Programming language is ……………….
(a) Flow chart
(b) Pseudo code
(c) C ++
(d) C
Answer:
(b) Pseudo code

Question 16.
Identify the statement which is wrong ……………….
(a) Programs must obey the grammar of the Programming language exactly.
(b) The punctuations in Programming language must be exactly.
(c) The Programming language is informal.
(d) Translator translates the programs into instructions.
Answer:
(c) The Programming language is informal.

Question 17.
Which one is used for converting programs into computer executable instructions?
(a) Converter
(b) Apps
(c) Translator
(d) exe files
Answer:
(c) Translator

Question 18.
Identify the correct statement from the following?
(a) Pseudo code can be executed by the computer directly.
(b) Pseudo code cannot be executed by the computer directly.
(c) Algorithm cannot be expressed in Pseudo code
(d) Algorithm are written only for python language
Answer:
(b) Pseudo code cannot be executed by the computer directly.

Question 19.
The notation which is not formal nor exact is ……………….
(a) Flow chart
(b) Pseudo code
(c) Compiler
(d) Translator
Answer:
(b) Pseudo code

Question 20.
Identify the correct statements.
(i) Pseudo code uses natural English for statements and conditions..
(ii) Pseudo code notation is formal one.
(iii) There is no need to follow the rules of the Programming language grammer
(iv) It must be rigorous and correct.
(a) (i) (ii) (iii)
(b) (ii) (iii) (iv)
(c) (i) (iii) (iv)
(d) (i) (ii) (iii) (iv)
Answer:
(c) (i) (iii) (iv)

Question 21.
Find the pair which is wrongly matched.
(a) Rectangular boxes – Statements
(b) Diamond boxes – Output
(c) Arrow – Control flow
(d) Parallelogram – Input
Answer:
(b) Diamond boxes – Output

Question 22.
Find the correct flow chart diagram from the following.

Answer:

Question 23.
The inputs and outputs are drawn using ……………….. boxes.
(a) rectangular
(b) diamond
(c) Parallelogram
(d) Oval
Answer:
(c) Parallelogram

Question 24.
The symbol used for representing the box to be executed next is ………………..
(a) ⇒
(b) ⇓
(c) ↓
(d) ++
Answer:
(c) ↓

Question 25.
The flow of control is represented in the flowchart by ………………..
(a) arrow
(b) dot
(c) box
(d) plus
Answer:
(a) arrow

Question 26.
In flow chart, rectangular boxes represents ………………..
(a) statements
(b) condition
(c) Input
(d) End
Answer:
(a) statements

Question 27.
A condition is contained in a diamond shaped box with ……………….. outgoing arrows.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 28.
A collection of boxes containing statements and conditions connected by arrow are ………………..
(a) compiler
(b) Flow chart
(c) Pseudo code
(d) Algorithm
Answer:
(b) Flow chart

Question 29.
Match the following:

(a) (i) -4 (ii) 0 – 1 (iii) – 2 (iv) – 3
(b) (i) – 1 (ii) – 2 (iii) – 3 (iv) – 4
(c) (i) – 4 (ii) – 2 (iii) – 1 (iv) – 3
(d) (i) – 3 (ii) – 2 (iii) – 1 (iv) – 4
Answer:
(a) (i) -4 (ii) 0 – 1 (iii) – 2 (iv) – 3

Question 30.
How many outgoing arrows are needed for rectangular boxes in flow chart?
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 31.
Statements composed of other statements are known as:
(a) Simple Statements
(b) Compound Statements
(c) Conditional
(d) Control flow
Answer:
(b) Compound Statements

Question 32.
Which one of the following is not a control flow statements?
(a) Sequential
(b) Assignment
(c) Iterative
(d) Alternative
Answer:
(b) Assignment

Question 33.
Which one of the following statement is used to alter the normal flow of control of the program?
(a) Assignments
(b) Control flow
(c) Compound
(d) both b & c
Answer:
(d) both b & c

Question 34.
Which one of the following statements are executed one after the other as written in the algorithm?
(a) Sequential
(b) Iterative
(c) Conditional
(d) Decisive
Answer:
(a) Sequential

Question 35.
Alternative statements analyses the problem into ……………….. cases.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 36.
Case analysis statement generalizes the statement into ……………….. cases.
(a) 2
(b) 3
(c) 5
(d) multiple
Answer:
(d) multiple

Question 37.
If _s^C is a ……………….. statement.
(a) Conditional
(b) Alternative
(c) Case Analysis
(d) Iterative
Answer:
(a) Conditional

Question 38.
Which one of the following process executes the same action repeatedly?
(a) Conditional
(b) Alternative
(c) Iterative
(d) None of these
Answer:
(c) Iterative

Question 39.
The iterative statement is commonly known as a ………………..
(a) loop
(b) Case Analysis
(c) Alternative
(d) Conditional
Answer:
(a) loop

Question 40.
Testing the loop condition and executing the loop body once is called ………………..
(a) alternative
(b) conditional
(c) Iteration
(d) Decomposition
Answer:
(c) Iteration

PART – 2
II. Short Answers

Question 1.
What is a Programming Language?
Answer:

1. A programming language is a notation for expressing algorithms to be executed by computers.
2. Programs must obey the grammar of the programming language exactly. Ex. C, C++, python.

Question 2.
What is a Pseudo code?
Answer:
Pseudo code is a mix of programming language like constructs and plain English. Algorithms expressed in Pseudo code are not intended to be executed by computers but for human readers to understand.

Question 3.
What is a Flow – chart?
Answer:
Flow chart is a diagrammatic notation for representing algorithms. A flow chart is a collection of boxes containing statements and conditions which are connected by arrows showing the order in which the boxes are to be executed.

Question 4.
What is a control flow statement. Classify it.
Answer:
Control flow statements are compound statements. They are used to alter the control flow of the process depending on the state of the process. They are classified as:

1. Sequential
2. Alternative
3. Iterative

Question 5.
What is meant by Decomposition?
Answer:
Decomposition is one of the elementary problem-solving techniques. An algorithm may be broken into parts, expressing only high level details. Then, each part may be refined into smaller parts, expressing finer details or each part may be abstracted as a function.

Question 6.
When a condition statement will be executed?
Answer:
It will be executed only when the condition statement is true.

Question 7.
Write the algorithm specification to find minimum of 2 numbers.
Answer:
Minimum (a, b)
– – inputs : a, b
– – outputs : result = a >1 b

Question 8.
Draw the flowchart for alternative control flow.
Answer:

PART – 3
III. Explain in Brief

Question 1.
Draw the diagram flowchart for integer division.
Answer:

Question 2.
Explain conditional statement.
Answer:
Conditional statement: Sometimes we need to execute a statement only if a condition is true and do nothing if the condition is false. This is equivalent to the alternative statement in which the else clause is empty. This variant of alternative statement is called a conditional statement. If C is a condition and S is a statement, then

is a statement, called a conditional statement, that describes the following action:

1. Test whether C is true or false.
2. If C is true then do S; otherwise do nothing.

The conditional control flow is depicted in the flowchart.

Question 3.
Write a note on refinement.
Answer:
After decomposing a problem into smaller subproblems, the next step is either to refine the subproblem or to abstract the subproblem.
1. Each subproblem can be expanded into more detailed steps. Each step can be further expanded to still finer steps, and so on. This is known as refinement.

2. We can also abstract the subproblem. We specify each subproblem by its input property and the input – output relation. While solving the main problem, we only need to know the specification of the subproblems. We do not need to know how the subproblems are solved.

Question 4.
Draw various symbols used in flowchart with their usage.
Answer:

Question 5.
Write a Algorithm to find minimum of 2 numbers.
Answer:

Question 6.

Answer:

Question 7.
Construct an iterative algorithm to compute the quotient and remainder after dividing an integer A by another integer B.
Answer:
divide (A, B)
– – inputs: A is an integer and B ≠ 0
– – Outputs: q and r such that A = q x B + r and 0 < r < B
q : = 0, A
While r ≥ B
q, r : = q + 1, r – B

PART – 4
IV. Explain in Detail

Question 1.
Write the disadvantages of flowcharts.
Answer:
Disadvantages of flowcharts:

1. Flowcharts are less compact than representation of algorithms in programming language or pseudo code.
2. They obscure the basic hierarchical structure of the algorithms.
3. Alternative statements and loops are disciplined control flow structures.

Question 2.
Explain the sequential statement in detail with the diagram.
Answer:
A sequential statement is composed of a sequence of statements. The statements in the sequence are executed one after another, in the same order as they are written in the algorithm, and the control flow is said to be sequential. Let S1 and S2 be statements. A sequential statement composed of S1 and S2 is written as
S1
S2
In order to execute the sequential statement, first do S1 and then do S2. The sequential statement given above can be represented in a flowchart. The arrow from S1 to S2 indicates that S1 is

Let the input property be P, and the input-output relation be Q, for a problem. If statement S solves the problem, it is written as

1. – – p
2. S
3. – – Q

If we decompose the problem into two components, we need to compose S as a sequence of two statements S1 and S2 such that the input-output relation of S1, say R, is the input property of S2.

1. – – P
2. S1
3. – – R
4. S2
5. – – Q

Question 3.
Write a note on case analysis.
Answer:
Alternative statement analyses the problem into two cases. Case analysis statement generalizes it to multiple cases. Case analysis splits the problem into an exhaustive set of disjoint cases. For each case, the problem is solved independently. If Cl, C2, and C3 are conditions, and S1, S2, S3 and S4 are statements, a 4 – case analysis statement has the form,

1. case C1
2. S1
3. case C2
4. S2
5. case C3
6. S3
7. else
8. S4

The conditions C1, C2, and C3 are evaluated in turn. For the first condition that evaluates to true, the corresponding sttement is executed, and the case analysis statement ends. If none of the conditions evaluates to true, then the default case S4 is executed.

1. The cases are exhaustive: at least one of the cases is true. If all conditions are false, the default case is true.
2. The cases are disjoint: only one of the cases is true. Though it is possible for more than one condition to be true, the case analysis always executes only one case, the first one that is true. If the three conditions are disjoint, then the four cases are (1) C1, (2) C2, (3) C3, (4) (not C1) and (not C2) and (not C3).

Question 4.
What is a function? Explain in detail.
Answer:
After an algorithmic problem is decomposed into subproblems, we can abstract the subproblems as functions. A function is like a sub-algorithm. Similar to an algorithm, a function is specified by the input property, and the desired input – output relation.

To use a function in the main algorithm, the user need to know only the specification of the function – the function name, the input property, and the input – output relation. The user must ensure that the inputs passed to the function will satisfy the specified property and can assume that the outputs from the function satisfy the input – output relation. Thus, users of the function need only to know what the function does, and not how it is done by the function. The function can be used a “black box” in solving other problems.

Ultimately, someone implements the function using an algorithm. However, users of the function need not know about the algorithm used to implement the function. It is hidden from the users. There is no need for the users to know how the function is implemented in order to use it.

An algorithm used to implement a function may maintain its own variables. These variables are local to the function in the sense that they are not visible to the user of the function. Consequently, the user has fewer variables to maintain in the main algorithm, reducing the clutter of the main algorithm.

Question 5.
Draw the flowchart for Eat breakfast.
Answer:

Question 6.
Explain Alternative statement.
Answer:
Alternative statement:
A condition is a phrase that describes a test of the state. If C is a condition and both S1 and S2 are statements, then

is a statement, called an alternative statement, that describes the following action:

1. Test whether C is true or false.
2. If C is true, then do S1; otherwise do S2.

In pseudo code, the two alternatives S1 and S2 are indicated by indenting them from the keywords if and else, respectively. Alternative control flow is depicted in the flowchart. Condition C has two outgoing arrows, labeled true and false. The true arrow points to the S1 box. The false arrow points to the S2 box. Arrows out of S1 and S2 point to the same box, the box after the alternative statement.

Conditional statement:
Sometimes we need to execute a statement only if a condition is true and do nothing if the condition is false. This is equivalent to the alternative statement in which the else clause is empty. This variant of alternative statement is called a conditional statement.

If C is a condition and S is a statement, then

1. Test whether C is true or false.
2. If C is true then do S; otherwise do nothing.

The conditional control flow is depicted in the flowchart.

Question 7.
Explain various symbols in Flow chart.
Answer:
Flowchart is a diagrammatic notation for representing algorithms. They show the control flow of algorithms using diagrams in a visual manner. In flowcharts, rectangular boxes represent simple statements, diamond-shaped box represent conditions, and arrows describe how the control flows during the execution of the algorithm. A flowchart is a collection of boxes containing statements and conditions which are connected by arrows showing the order in which.the boxes are to be executed.

1. A statement is contained in a rectangular box with a single outgoing arrow,which points to the box to be executed next.
2. A condition is contained in a diamond shaped box with two outgoing arrows, labeled true and false. The true arrow points to the box to be executed next if the condition is true, and the false arrow points to the box to be executed next if the condition is false.
3. Parallelogram boxes represent inputs given and outputs produced.
4. Special boxes marked Start and the End are used to indicate the start and the end of an execution:

The flowchart of an algorithm to compute the quotient and remainder after dividing an integer A by another integer B is shown below, illustrating the different boxes such as input, output, condition, and assignment, and the control flow between the boxes.

Flowcharts also have disadvantages.

1. Flowcharts are less compact than representation of algorithms in programming language or pseudo code.
2. They obscure the basic hierarchical structure of the algorithms.
3. Alternative statements and loops are disciplined control flow structures. Flowcharts do not . restrict us to disciplined control flow structures.

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

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Samacheer Kalvi 12th Chemistry Chapter 8 Ionic Equilibrium Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Multiple Choice Questions

I. Choose the correct answer.

12th Chemistry Ionic Equilibrium Question 1.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 x 10^-4 mol L^-1 solubility product of Ag₂C₂O₄ is ………………
(a) 2.42 x 10^-8 mol³ L^-3
(b) 2.66 x 10^-12 12 mol³ L^-3
(c) 45 x 10^-11 mol³ L^-3
(d) 5.619 x 10^-12 mol³ L^-3
Answer:
(d) 5.619 x 10^-12 mol³ L^-3
Ag₂C₂O₄ 2Ag⁺ + C₂ O₄^2-
[Ag⁺] = 2.24 x 10^-4 mol L^-1

= 1.12 x 10^-4 mol L^-1
K_sp = [Ag⁺]² [C₂O₄^2-]
=(2.24 x 10^-4 mol L^-1)² (1.12 x 10^-4 mol L^-1)
=5.619 x 10^-12 mol³ L^-3

12th Chemistry Chapter 8 Book Back Answers Question 2.
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.
(i) 60 mL \(\frac { M }{ 10 }\) HCI + 40 mL \(\frac { M }{ 10 }\) NaOH
(ii) 55 mL \(\frac { M }{ 10 }\) HCl + 45 mL \(\frac { M }{ 10 }\) NaOH
(iii) 75 mL \(\frac { M }{ 5 }\) HCI +25 mL \(\frac { M }{ 5 }\) MNaOH
(iv) 100 mL \(\frac { M }{ 10 }\) HCI+ 100 mL \(\frac { M }{ 10 }\) NaOH

pH of which one of them wilt be equal to 1?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Answer:
(d) (iii) 75 mL \(\frac { M }{ 5 }\) HCI + 25 mL \(\frac { M }{ 5 }\) NaOH
No of moles of HCl = 0.2 x 75 x 10^-3 = 15 x 10^-3
No of moles of NaOH = 0.2 x 25 x 10^-3 = 5 x 1o^-3
No of moles of HCl after mixing = 15 x 10^-3 – 5 x 10^-3
∴ Concentration of HCl

for (iii) solution, pH of 0.1 M HCI = – 1og₁₀ (0.1) = 1.

12th Chemistry 8th Lesson Book Back Answers Question 3.
The solubility of BaSO₄ in water is 2.42 x 10^-3 gL^-1 at 298K. The value of its solubility product
(K_sp) will be …………………..
(Given molar mass of BaSO₄ = 233g mol^-1)
(a) 1.08 x 10^-14 mol²L²
(b) 1.08 x 10^-12 mol²L²
(c) 1.08 x 10^-10 mol² L²
(d) 1.08 x 10^-8 mol²L^-2
Answer:
(c) 1.08 x 10^-10 mol² L²
BaSO₄ \(\rightleftharpoons\) Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²

Ionic Equilibrium Questions And Answers Pdf Question 4.
pH of a saturated solution of Ca(OH)₂ is 9. The Solubility product (K) of Ca(OH)₂ ………………..
(a) 0.5 x 10^-15
(b) 0.25 x 10^-10
(c) 0.125 x 10^-15
(d) 0.5 x 10^-10
Answer:
(a) 0.5 x 10^-15
Ca(OH)₂ \(\rightleftharpoons\) Ca²⁺ + 2OH^–
Given that pH = 9
pOH = 14 – 9 = 5
[p0K = – 1og₁₀[OH^–]]
[OH^–] = 10^-pOH
[OH^–] =10^-5M
K_sp = [Ca²⁺] [OH^–]²

=0.5

Class 12 Ionic Equilibrium Question 5.
Conjugate base for bronsted acids H₂O and HF are ………………
(a) OH^– and H₂FH⁺, respectively
(b) H₃O⁺ and F^–, respectively
(c) OH^– and F^–, respectively
(d) H₃O⁺ and H₂F⁺, respectively
Answer:
(c) OH^– and F^–, respectively

∴ Conjugate bases are OH^– and F^– respectively

Ionic Equilibrium Notes Pdf Question 6.
Which will make basic buffer?
(a) 50 mL of 0.1M NaOH + 25mL of 01M CH₃COOH
(b) 100 mL of 0.1M CH₃COOH + 100 mL of 0.1M NH₄OH
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH₄OH
(d) 100 mL of 0.1M HCI + 100 mL of O.1 M NaOH
Answer:
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH₄OH
Basic buffer is the solution which has weak base and its salt

Samacheer Kalvi Guru 12th Chemistry Question 7.
Which of the following fluro – compounds is most likely to behave as a Lewis base?
(a) BF₃
(b) PF₃
(c) CF₄
(d) SiF₄
Answer:
(b) PF₃
BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CF₄ → neutral → neither lewis acid nor base
SiF₄ → neutral → neither lewis acid nor base

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.
Which of these is not likely to act as lewis base?
(a) BF₃
(b) PF₃
(c) CO
(d) F^–
Answer:
(a) BF₃
BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
F^– → unshared pair of electron → lewis base

Ionic Equilibrium Notes Question 9.
What is the decreasing order of strength of bases?
OH^–, NH₂^–, H – C = C^– and CH₃ – CH₂^–
(a) OH^– > NH₂^– > H – C = C > CH₃ – CH₂^–
(b) NH₂^– > OH^– > CH₃ – CH₂^– > H – C = C^–
(c) CH₃ – CH₂^–, > NH₂^– > H – C = C^– > OH^–
(d) OH^– > H – C = C > CH₃ – CH₂^– > NH₂^–
Answer:
(c) CH₃ – CH₂^–, > NH₂^– > H – C = C^– > OH^–
Acid strength decreases in the order
HOH > CH = CH > NH₃ > CH₃CH₃
Its conjucate bases arc in the reverse order
CH₃ – CH₂^– > NH₂^– > H – C = C > OH^–

Ionic Equilibrium Question 10.
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
(a) acidic, acidic, basic
(b) basic, acidic, basic
(c) basic, neutral, basic
(d) none of these
Answer:
(b) basic, acidic, basic

Class 12 Chemistry Chapter 8 Notes Question 11.
The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅NH) in a 0.10M aqueous pyridine solution (Kb for C₅H₅N = 1.7 x 10^-9) iS ……………..
(a) 0.006%
(b) 0.013%
(c) 0.77%
(d) 1.6%
Answer:
(b) 0.013%

Percentage of dissociation
= \(\sqrt { 1.7 }\) x 10^-4 x 100 = 1.3 x 10^-2 = 0.013%

Question 12.
Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
(a) 37 x 10^-2
(b) 10^-6
(c) 0.111
(d) none of these
Answer:
(a) 3.7 x 10^-2
pH = – log₁₀ [H⁺]
[H⁺] = 10^-pH
Let the volume be x mL
V₁M₁ + V₂M₂ + V₃M₃ = VM
x mL of 10^-1M + x mL of 10^-2M + x mL of 10^-3 M
= 3 x mL of [H⁺]
= 3 x mL of [H⁺]
[H⁺] =

= 0.037 = 3.7 x 10^-2

Question 13.
The solubility of AgCl (s) with solubility product 1.6 x 10^-10 in 0. 1 M NaCl solution would be ………….
(a) 1.26 x 10^-5 M
(b) 1.6 x 10^-9 M
(c) 1.6 x 10^-11 M
(d) Zero
Answer:
(b) 1.6 x 10^-9 M
AgCl (s) \(\rightleftharpoons\) Ag⁺(aq) + Cl^–(aq)

K_sp = 1.6 x 10^-10
K_sp = [Ag⁺][Cl^–]
K = (s) (s+0.1)
0.1 >>s

Question 14.
If the solubility product of lead iodide is 3.2 x 10^-8, its solubility will be …………..
(a) 2 x 10^-3M
(b) 4 x 10^-4 M
(c) l.6 x 10^-5 M
(d) 1.8 x 10^-5 M
Answer:
(a) 2 x 10^-3M
PbI₂ (s) → Pb²⁺ (aq) + 2I^– (aq)
K_sp = (s) (2s)²
3.2 x 10^-8 = 4s³

Question 15.
Using Gibb’s free energy change, ∆G⁰ = 57.34 KJ mol^-1, for the reaction, X₂Y_(g) \(\rightleftharpoons\) 2X⁺ + Y^2-(aq), calculate the solubility product of X₂Y in water at 300K (R = 8.3 J K^-1 Mol^-1) ……………….
(a) 10^-10
(b) 10^-12
(c) 10^-14
(d) can not be calculated from the given data
Answer:
(a) 10^-10
57.34 KJ mol^-1 = – 2.303 x 8.3 JK^-1 mol^-1 x 300K log K_sp

log₁₀K_sp = -10
∴ K_sp = 10^-10
∆G⁰ = – 2.303 RT log K_eq
X₂Y_(s) \(\rightleftharpoons\) 2X⁺(aq) +Y^2-(aq)

K_eq = [x⁺]²[Y^2-] ( X₂Y_(s) = 1)
K_eq = K
Question 16.
MY and NY₃, are insoluble salts and have the same K_sp values of 6.2 x 10^-13 at room temperature. Which statement would be true with regard to MY and NY₃?
(a) The salts MY and NY₃ are more soluble in O.5 M KY than in pure water
(b) The addition of the salt of KY to the suspension of MY and NY₃ will have no effect on
(c) The molar solubities of MY and NY₃ in water are identical
(d) The molar solubility of MY in water is less than that of NY₃
Answer:
(d) The molar solubility of MY in water is less than that of NY₃
Addition of salt KY (having a common ion Y^–) decreases the solubility of MY and NY₃ due to common ion effect. Option (a) and (b) are wrong.
For salt MY, MY \(\rightleftharpoons\) M⁺ + Y^–
K_sp = (s) (s)
6.2 x 10^-13 = s²

The molar solubility of MY in water is less than of NY₃

Question 17.
What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
(a) 2.0
(b) 3
(c) 7.0
(d) 12.65
Answer:
(d) 12.65
x ml of 0.1 m NaOH + x ml of 0.01 M HCI
No. of moles of NaOH = 0.1 x x x 10^-3 = 0.l x x 10^-3
No. of moles of HCl = 0.01 x x x 10^-3 = 0.01 x x 10^-3
No. of moles of NaOH after mixing = 0.1x x 10^-3 – 0.01x  x 10^-3
= 0.09x x 10^-3
Concentration of NaOH =

[OH^–] = 0.045
p^OH = – log (4.5 x 10^-2)
= 2 – log 4.5
= 2 – 0.65 = 1.35
pH = 14 – 1.35 = 12.65

Question 18.
The dissociation constant of a weak acid is 1 x 10^-3 . In order to prepare a buffer solution with a pH =4, the [Acid] / [Salt] ratio should be ………………..
(a) 4:3
(b) 3:4
(c) 10:1
(d) 1:10
Answer:
(d) 1:10
K_a = 1 x 10^-3 ; pH = 4

Question 19.
The pH of 10^-5 M KOH solution will be …………..
(a) 9
(b) 5
(c)19
(d) none of these
Answer:
(a) 9

[OH^–] = 10^-5M.
pH = 14 – pOH .
pH = 14 – ( – log [OH^–])
= 14 + log [OH^–] = 14 + log 10^-5
= 14 – 5 = 9

Question 20.
H₂PO₄^– the conjugate base of …………….
(a) PO₄
(b) P₂O₅
(c) H₃PO₄
(d) HPO₄^2-
Answer:
(c) H₃PO₄

H₂PO₄ is the conjugate base of H₃PO₄

Question 21.
Which of the following can act as lowery – Bronsted acid well as base?
(a) HCl
(b) SO₄^2-
(c) HPO₄^2-
(d) Br^–
Answer:
(c) HPO₄^2-
HPO₄^2- can have the ability to accept a proton to form H₂PO₄.
It can also have the ability to donate a proton to form PO₄^-3.

Question 22.
The pH of an aqueous solution is Zero. The solution is ……………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(b) strongly acidic
pH = – log₁₀[H⁺]
[H⁺] =10^-pH
= 10⁰ = 1
[H⁺] = 1 M
The, solution is strongly acidic

Question 23.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by ………………

Answer:
According to Henderson equation

According to Henderson equation

Question 24.
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?

Answer:

Question 25.
Dissociation constant of NH₄OH is 1.8 x 10^-5 the hydrolysis constant of NH₄Cl would be …………….
(a) 1.8 x 10^-19
(b) 5.55 x 10^-10
(c) 5.55 x 10^-5
(d) 1.80 x 10^-5
Answer:
(b) 5.55 x 10¹⁰

II. Answer the following questions.

Question 1.
What are lewis acids and bases? Give two example for each.
Answer:
1. Lewis acids:

• Lewis acid is a species that accepts an electron pair.
• Lewis acid is a positive ion (or) an electron deficient molecule.
• Example, Fe²⁺, CO₂, BF₃, SiF₄ etc…

2. Lewis bases:

• Lewis base is a species that donates an electron pair.
• Lewis base is an anion (or) neutral molecule with atleast one lone pair of electrons.
• Example, NH₃, F^–, CH₂ = CH₂, CaO etc….

Question 2.
Discuss the Lowry – Bronsted concept of acids and bases.
Answer:
According to Lowry – Bronsted concept, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance. When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCl behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCl + H₂O \(\rightleftharpoons\) H₃O⁺ + Cl^–

When ammonia is dissolved in water, it accepts a proton from water. In. this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ \(\rightleftharpoons\) NH₄⁺ + OH^–
Let us consider the reverse reaction in the following equilibrium

H₃O⁺ donates a proton to Cl^– to form HCI i.e., the products also behave as acid and base. In general, Lowry – Bronsted (acid – base) reaction is represented as
Acid₁ + Base₂ \(\rightleftharpoons\) Acid₂ + Base₁
The species that remains after the donation of a proton is a base (Base₁)and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs. Conjugate acid – base pair

HCl and Cl^–, H₂O and H₃O are two conjugate acid – base pairs. i.e., Cl_– is the conjugate base of the acid HCl (or) HCl is conjugate acid of Cl^– Similarly H₃O is the conjugate acid of H₂O. Limitations of Lowry – Bronsted theory. Substances like BF₃ , AICl₃ etc., that do not donate protons are known to behave as acids.

Question 3.
Indentify the conjugate acid base pair for the following reaction in aqueous solution.

1. HS^– (aq) + HF \(\rightleftharpoons\) F^– (aq) + H₂S (aq)
2. HPO₄^2- + SO₃^2- \(\rightleftharpoons\) PO₄^3- + HSO₃^–
3. NH₄⁺ + CO₃^2- \(\rightleftharpoons\) NH₃ + HCO₃^–

Answer:
1.

• HF and F^– , HS^– and H₂S are two conjugate acid – base pairs.
• F^– is the conjugate base of the acid HF (or) HF is the conjugate acid of F^–
• H₂S is the conjugate acid of HS^– (or) HS^– is the conjugate base of H₂S.

2.

• HPO₄^2- and PO₄^3-, SO₃^2- and HSO₃^– are two conjugate acid – base pairs.
.PO₄^3- is the conjugate base of the acid HPO₄^2- (or) HPO₄^2- is the conjugate acid of PO₄.
•HSO₃^– is the conjugate acid of SO₃^2- (or) SO₃^2- is the conjugate base of HSO₃^–.

3.

• NH⁺ and NH3, CO3^2- and HCO3^– are two conjugate acid – base pairs.
• HCO₃^– is the conjugate of acid CO₃^2- (or) CO₃^2- is the conjugate bases of HCO₃^–.
• NH₃ is the conjugate base of NH₄⁺ (or) NH₄⁺ is the conjugate acid of NH₃.

Question 4.
Account for the acidic nature of HCIO4. In terms of Bronsted – Lowry theory, identify its conjugate base.
Answer:
HClO₄ \(\rightleftharpoons\) H⁺ + ClO₄^–
1. According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

2. Let us consider the stabilities of the conjugate bases ClO₄^– , ClO₃^–, CIO₂^– and ClO^– formed from these acid HClO₄, HClO₃, HCIO₂, HOCI respectively.

These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

3. The charge stabilization mercases in the order, ClO^– < ClO₂^– < ClO₃^– < ClO₄^– .

This means ClO₄^– will have maximum stability and therefore will have minimum attraction for W. Thus CIO₄^– will be weakest base and its conjugate acid HCIO₄ is the strongest acid.

4. CIO₄^– is the conjugate base of the acid HClO₄.

Question 5.
When aqueous ammonia is added to CuSO₄ solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, [Cu(H₂O)₆]²⁺(aq) + 4NH₃ (aq) \(\rightleftharpoons\) [Cu(NH₃)₄]²⁺ (aq), among H₂O and NH₃ which is stronger Lewis base.
Answer:
Copper (II) sulphate solution, for example contains the blue hexaaqua copper (II) complex ion. In the first stage of the reaction, the ammonia acts as a Bronsted – Lowry base. With a small amount of ammonia solution, hydrogen ions are pulled off two water molecules in the hexaaqua ion. This produces a neutral complex, one carrying no charge.

If you remove two positively charged hydrogen ions from a 2+ ion, then obviously there isn’t going to be any charge left on the ion. Because of the lack of charge, the neutral complex isn’t soluble in water and so you get a pale blue precipitate. [Cu(H₂O)⁶]²⁺ + 2NH₃ [Cu(H₂O)₄OH] + 2NH₄⁺

This precipitate is often written as Cu(OH)₂ and called copper (II) hydroxide. The reaction is reversible because ammonia is only a weak base. That precipitate dissolves if you add an excess of ammonia solution, giving a deep blue solution. The ammonia replaces four of the water molecules around the copper to give tetramminc diaqua copper (II) ions. The ammonia
uses its lone pair to form a coordinate covalent bond with the copper. It is acting as an electron pair donor – a Lewis base.

Question 6.
The concentration of hydroxide ion in a water sample is found to be 2.5 x 10^-6 M. Identify the nature of the solution.
Answer:
The concentration of OH ion in a water sample is found to be 2.5 x 10 M
pOH = – log₁₀ [OH^– ]
pOH = – 1og₁₀ [2.5 x 10^-6]
= – log₁₀ [2.5] – log₁₀ [10^-6]
= – 0.3979 – ( – 6)
= – 0.3979 + 6
pOH = 5.6

We know that,
pH + pOH = 14
pH + 5.6 = 14
pH = 14 – 5.6
pH = 8.4
pH = 8.4, shows the nature of the solution is basic.

Question 7.
A lab assistant prepared a solution by adding a calculated quantity of HCl gas 25°C to get a solution with [H₃O⁺] = 4 x 10⁵ M. Is the solution neutral (or) acidic (or) basic.
Answer:
[H₃O⁺] = 4 x M
pH = – log₁₀ [H₃O⁺]
pH = – 1og₁₀[4 x 10⁵]
pH = – log₁₀ [4] – log₁₀ [10^-5]
pH = – 0.6020 – ( – 5) = – 0.6020 + 5
pH = 4.398
Therefore, the solution is acidic.

Question 8.
Calculate the pH of 0.04 M HNO₃ Solution.
Answer:
Concentration of HNO₃ = 0.04M
[H₃O⁺] = 0.04 mol dm^-3
pH = – 1og[H₃O⁺]
= – log (0.04)
= – log(4 x 10^-2)
= 2 – log4 = 2 – 0.6021
= 1.3979 = 1.40

Question 9.
Define solubility product.
Answer:
Solubility product:
It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric coefficient in a balanced equilibrium equation.

Question 10.
Define ionic product of water. Give its value at room temperature.
Answer:
1. The product of concentration of H⁺ and OH^– ions in water at a particular temperature is known as ionic product.
2. The ionic product of water at room temperature (25°C) is,
K_w = [H⁺] [OH⁺] (or)
K_w= [H₃O⁺] [OH⁺]
K_w =(1 x 10^-7) (1 x 10^-7)
K_w= 1 x 10^-14 mol² dm^-6

Question 11.
Explain common ion effect with an example.
Answer:
Common ion Effect:
When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists.
CH₃COOH (aq) \(\rightleftharpoons\) H⁺(aq)+ CH₃COO^– (aq)

However, the added salt, sodium acetate, completely dissociates to produce Na⁺ and CH₃COO ion.
CH₃COONa (aq) → Na⁺ (aq) + CH₃COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier’s priñciple that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH₃COO^– ions combines with H ions to produce much more unionized CH₃COOH i.e.,

the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 12.
Derive an expression for Ostwald’s dilution law.
Answer:
Ostwald’s dflution law:
It relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c). Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,
CH₃COOH \(\rightleftharpoons\) CH₃COO^– + H⁺
The dissociation constant of acetic acid is,


Substituting the equilibrium concentration in equation

We know that weak acid dissociates only to a very small extent compared to one, a is so small.
equation (1) becomes,

Similarly, for a weak base,

The concentration of H can be calculated using the K_a value as below,

Substituting a value in equation (2),

For weak base

Question 13.
Define pH.
Answer:
pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.
pH = – log₁₀ [H₃O] (or) pH = – log₁₀ [H⁺]

Question 14.
Calculate the pH of 1.5 x 10^-3 M solution of Ba(OH)₂
Answer:

[OH^–] = 3 x 10³M.
[pH + pOH = 14]
pH = 14 – pOH
pH = 14 – ( – log [OH^–])
= 14 + log [OH^–]
= 14 + log (3 x 10^-3)
= 14 + log 3 + log 10^-3
= 11 + 0.4771
pH = 11.48

Question 15.
50 ml of 0.05 M HNO₃ is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.
Solution.
Number of moles of HNO₃ = 0.05 x 50 x = 2.5 x 10^-3
Number of moles of KOH = 0.025 x 50 x 10^-3 = 1.25 x 10^-3
Number of moles of HNO₃ after mixing = 2.5 x 10^-3 – 1.5 x 10^-3
= 1.25 x 10^-3

After mixing, total volume = 100 ml = 100 x 10^-3 L

pH = – log [H⁺]
pH = – log (1.25 x 10^-2) = 2 – 0.0969
= 1.9031

Question 16.
The K_a value for HCN is 10^-9. What is the pH of 0.4 M HCN solution?
Answer:
K_a =10^-9
c = O.4M
pH = – log [H⁺]

∴ pH = – log(2 x 10^-5)
= – log 2 – log (10^-5)
= – 0.3010 + 5
pH = 4.699

Question17.
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that.
K_a = K_b = 1.8 x 10^-5
Solution.

= 0.7453 x 10^-2
pH = \(\frac { 1 }{ 2 }\) pK_w + \(\frac { 1 }{ 2 }\) pK_a – \(\frac { 1 }{ 2 }\) pK_b
Given that K_a = K_b = 1.8 x 10^-5
if K_a = K_b, then. pK_a = pK_b
pH = \(\frac { 1 }{ 2 }\) pK_w = \(\frac { 1 }{ 2 }\) (14) = 7

Question 18.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
Answer:
Let us consider the reactions between a strong acid, HCl, and a weak base, NH₄OH, to produce a salt, NH₄Cl, and water.
HCl (aq) + NH₄OH (aq) \(\rightleftharpoons\) NH₄Cl (aq) + H₂O (I)
NH₄CI(aq) → NH₄⁺ + Cl^– (aq)

NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^– from water to produce unionised NH₄OH shown below.
NH₄⁺ (aq) + H₂O (1) \(\rightleftharpoons\) NH₄OH (aq) + H⁺(aq)

There is no such tendency shown by Ct and therefore [H⁺] > [OH^–] the solution is acidic and the pH is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_a and K_b as
K_h.K_b = K_w
Let us calculate the Kb value in terms of degree of hydrolysis (h) and the concentration of salt
K_h = h² C and

Question 19.
Solubility product of Ag₂CrO₄ is 1 x 10^-12. What is the solubility of Ag₂CrO₄ in 0.01 M AgNO₃ solution?
Answer:
Solubility product of Ag₂CrO₄,
K_sp = 1 x 10^-2

K_sp = [Ag⁺]² [ CrO₄^2-]
[Ag⁺] = 2s +0.01
0.01 >> 2s
[Ag⁺]  = 0.01M
[CrO₄^-2] = s
K_sp = (0.01)². (s)

Question 20.
Write the expression for the solubility product of Ca₃(PO₄)₂
Answer:
Ca₃(PO₄)₂ (s) \(\rightleftharpoons\) 3Ca²⁺ (3s) + 2PO₄^3- (2s)
Solubility of Ca₃(PO₄)₂ is,
K_sp = [Ca²⁺]³ . [PO₄^3-]²
K_sp = (3s)³ . (2s)²
K_sp= (27 s³) . (4s²)
K_sp = 108s⁵.

Question 21.
A saturated solution, prepared by dissolving CaF₂(s) in water, has [Ca²⁺] = 3.3 x 10^-4 M. What is the K_spof CaF₂?
Answer:
CaF_{2 (s)} \(\rightleftharpoons\) Ca²⁺_(aq) + 2F_–(aq)
[F_–] = 2 [Ca²⁺] = 2 x 33 x 10^-4 M
= 6.6 x 10^-4 M
= [Ca²⁺] [F^–]²
= (3.3 x 10^-4) (6.6 x 10^-4)²
= 1.44 x 10^-10

Question 22.
K_sp of AgCI is 1.8 x 10^-10. Calculate molar solubility in 1 M AgNO₃
Answer:
AgCI(s) Ag⁺(aq) + Cl^–(aq)
x = solubility of AgCI in 1M AgNO₃

[Cl^–] = x
K_sp = [Ag⁺] [Cl^–]
1.8 x 10^-10 = (1) (x)
x = 1.8 x 10^-10M

Question 23.
A particular saturated solution of silver chromate Ag₂CrO₄ has [Ag⁺] = 5 x 10^-5 and [CrO₄]^2- = 4.4 x 10 M. What is the value of for Ag₂CrO₄?
Answer:
Ag₂CrO₄ (s) \(\rightleftharpoons\) 2Ag⁺(aq) + CrO₄^2-(aq)
K_sp = [Ag⁺]² [CrO₄^2-]
= (5 x 10^-5)² (4.4 x 10^-4)
= 1.1 x 10^-12

Question 24.
Write the expression for the solubility product of Hg₂CI₂.
Answer:

Question 25.
K_sp of Ag₂CrO₄ is 1.1 x 10^-12 What is solubility of Ag₂CrO₄ in 0.1M K₂CrO₄.
Answer:

x is the solubility of Ag₂CrO₄ in 0.1 M K₂CrO₄

Question 26.
Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO₃)₂ and 0.100 L of 0.2 M NaCl are mixed? (PbCI₂) = 1.2 x 10^-5.
Answer:
When two or more solution are mixed, the resulting concentrations are differnet from the original.
Total volume 0.250L

Precipitation of PbCI₂ (s) occurs if [Pb²⁺][Cl^–]² > K_sp
[Pb²⁺][Cl^–]² = (0.06)(0.08)²
= 3.84 x 10^-4
Since ionic product [Pb²⁺][Cl^–]² > K_sp PbCl₂ is precipitated.

Question 27.
of Al(OH)₃ is 1 x 10^-15 M. At what pH does 1.0 x 10^-13 M AI³⁺ precipitate on the addition of buffer of NH₄CI and NH₄OH solution?
Answer:
Al(OH)₃ Al³⁺ (aq) + 3OH^– (aq)
K_sp = [Al³⁺] [OH^–]³
Al(OH)₃ precipitates when
[Al³⁺] [OH^–]³ > K_sp
(1 x 10^-3)[OH^–]³ > K_sp
[OH^–]³ > 1 x 10^-12
[OH^–] > 1 x 10^-4M
[OH^–] = l x 10^-4 M
pOH = – 1og₁₀[OH^–] = – log (1 x 10^-4) = 4
pH = 14 – 4 = 10
Thus, Al (OH)₃ precipitates at a pH of 10

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Evaluate Yourself

Question 1.
Classify the following as acid (or) base using Arrhenius concept

1. HNO₃
2. Ba(OH)₂
3. HlPO₄
4. CH₃COOH

Answer:
1. HNO₃:
Nitric acid, dissociates to give hydrogen ions in water.
HNO₃ is acid.

2. Ba(OH)₂:
Barium hydroxide, dissociates to give hydroxyl ions in water.
Ba(OH)₂ is base.

3. H₃PO₄:
Orthophosphoric acid, dissociates to give hydrogen ions in water.
H₃PO₄ is acid.

4. CH₃COOH:
Acetic acid, dissociates to give hydrogen ions in water.
CH₃COOH is acid.

Question 2.
Write a balanced equation for the dissociation of the following in water and identify the conjugate acid base pairs.

1. NH₄
2. H₂SO₄
3. CH₃COOH.

Answer:
1. NH₄ ⁺ Conjugate acid-base pair

NH₄⁺ and NH₃, H₂O and H₃O⁺ are two conjugate acid – base pairs.

H₂SO₄ and CH₃COO^–, H₂O and H₃O⁺ are two conjugate acid-base pairs.

2. H₂SO₄:

H₂SO₄ and HSO₄^–, H₂O and H₃O⁺ are two conjugate acid-base pairs.

3. CH₃COOH:

CH₃COOH and CH₃COO^–, H₂O and H₃O⁺ are two conjugate acid-base pairs.

Question 3.
Identify the Lewis acid and the Lewis base in the following reactions.

Answer:
(i). CaO + CO₂ → CaCO₃

• CaO – Lewis base – All metals oxides are Lewis bases
• CO₂ – Lewis acid – CO₂ contains a polar double bond.

• CH₃ – O – CH₃ – Lewis base – Electron rich species
• AlCl₃ – Lewis acid – AICI₃ is electron deficient molecule.

Question 4.
H₃BO₃ accepts hydroxide ion from water as shown below
Answer:
H₃BO₃ (aq) + H₂O(l) = B(OH)₄^– + H⁺
Predict the nature of H₃BO₃ using Lewis concept. Boric acid is also called as hydrogen borate or orthoboric acid. It is a weak mono basic Lewis acid of boron and it is written as B(OH)₃. It accepts hydroxyl (OH^–) ion from water. It does not dissociate to give hydronium (H₃O⁺) ion rather forms metaborate ion and this ions in turn give H₃O ion.
B(OH)₃ + H₂O [B(OH)₄]^– + H₃O⁺ Hence it is considered as weak acid.

Question 5.
At a particular temperature, the K_w of a neutral solution was equal to 4 x 10^-14. Calculate the concentration of [H₃O⁺] and [OH^–].
Answer:
Given solution is neutral
[H₃O⁺] = [OH^–]
Let [H₃O⁺] = x ; then [OH^–] = x
K_w = [H₃O⁺] [OH^–]
4 x 10^-14 = x . x
x² = 4 x 10^-14

Question 6.

1. Calculate pH of 10^-8 M H₂SO₄
2. Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4
3. Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCI with 50 ml 0.1 M NaOH

Answer:
1.

In this case the concentration of H₂SO₄ is very low and hence [H₃O] from water cannot be neglected
[H₃O⁺] = 2 x 10^-8 (from H₂SO₄) + 10^-7 (from water)
= 10^-8(2+ 10)
= 12 x 10^-8 = 1.2 x 10^-7
pH = – log₁₀[H₃O⁺]
= – log₁₀( 1.2 x 10^-7)
= 7 – log₁₀1.2
= 7 – 0.0791 = 6.9209

2.
pH of the solution = 5.4
[H₃O⁺] = antilog of (- pH)
= antilog of (- 5.4)
= antilog of (-6 + 0.6) = \(\overline{6} .6\)
= 3.981 x 10^-6
i.e., 3.98 x 10^-6 mol dm^-3

3.
No. of moles of HCl = 0.2 x 50 x 10^-3 = 10 x 10^-3
No. of moles of NaOH =0.1 x 50 x 10^-3 = 5 x 10^-3
No. of moles of HCl after mixing = 10 x 10^-3 – 5 x 10^-3
= 5 x 10^-3
after mixing total volume = 100 mL

[H₃O⁺] = 5 x 10^-2 M
pH = – log ( 5 x 10^-2)
= 2 – log 5
= 2 – 0.6990
= 1.30

Question 7.
K_b for NH₄OH is 1.8 x 10^-5 Calculate the percentage of ionisation of 0.06 M ammonium hydroxide solution.
Answer:

Question 8.
1. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.

2. Calculate the pH of a buffer solution consisting of 0.4M CH₃COOH and 0.4 M CH₃COONa . What is the change in the pH after adding 0.01 mol of HCI to 500m1 of the above buffer solution.

Assume that the addition of HCI causes negligible change In the volume. Given: (K = 1.8 x 105).
Answer:
1. Dissociation of buffer components
NH₄OH (aq) \(\rightleftharpoons\) NH₄⁺ (aq) + OH^– (aq)
NH₄CI → NH₄⁺ + Cl^–
Addition of OH^–
The added H⁺ ions are neutralized by NH₄OH and there is no appreciable decrease in pH.
NH₄OH(aq) + H⁺ \(\rightleftharpoons\) NH₄⁺(aq) + H₂O (1)
Addition of
NH₄^– (aq) + OH^– (aq) → NH₄OH (aq)
The added OH ions react with NH₄ to produce unionized NH₄OH . Since NH₄OH is a weak base, there is no appreciable increase in pH.

2. pH of buffer

Addition of 0.01 mol HCI to 500 ml of buffer
Added [H⁺]

∴ pH = – log (1.8 x 10^-5) = 4.74
Addition of 0.01 mol HCl to 500ml of buffer

Question 9.
1. How can you prepare a buffer solution of pH9. You are provided with 0.1 M NH₄OH solution and ammonium chloride crystals. (Given: pKb for NH₄OH is 4.7 at 25°C)

2. What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid. (Given: pK_a for formic acid is 3.75.)
Answer:
1.

We know that pH + pOH = 14
9 + pOH = 14
= pOH = 14 – 9 = 5


[NH₄Cl] = 0.1 M x 1.995
= 0. 1995 M
=0.2 M
Amount of NH₄CI required to prepare 1 litre 0.2 M solution = Strength of NH₄CI x molar
mass of NH₄CI
= 0.2 x 535
= 10.70 g
10.70 g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2 M solution. On mixing equal volume of the given NH₄OH solution and the prepared NH₄CI solution will give a buffer solution with required pH value (pH = 9).

2.

[Sodium formate] = number of moles of HCOONa
= 0.6 x V x 10^-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10^-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10^-3
= 80 x 10^-3
4 = 3.75 + log \(\frac { 0.6V }{ 80 }\)
0.25 = log \(\frac { 0.6V }{ 80 }\)
antilog of 0.25 = \(\frac { 0.6V }{ 80 }\)
0.6V = 1.778 x 80
= 1.78 x 80
= 142.4
V = \(\frac { 142.4 mL }{ 0.6 }\) = 237.33 mL

Question 10.
Calculate the

1. hydrolysis constant
2. degree of hydrolysis and
3. pH of 0.05M sodium carbonate solution pK_a for HCO₃^– is 10.26.

Answer:
1. Hydrolysis constant

Given
K_w = 1 x 10^-14
c = 0.05 M
PK_a = 10.26
K_a = – log K_a
K_a = antilog of (- pK_a)
K_a = antilog of (- 10.26)
K_a = 5.49 x 10^-11

h = 6.034 x 10^-2

2. Degree of hydrolysis

3.

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Textbook Example problems solved

Question 1.
Identify the Lewis acid and the Lewis base in the following reactions.
Cr³⁺ + 6H₂O → [Cr(H₂O)₆]³⁺
In the hydration of ion, each of six water molecules donates a pair of electron to Cr³⁺ to form the hydrated cation, hexaaquachromium (III) ion, thus, the lewis acid is Cr and the Lewis base H₂O.

Question 2.
Calculate the concentration of OH^– in a fruit juice which contains 2 x 10 M, H₃O⁺ Ion. Identify the nature of the solution.
Answer:
Given that H₃O⁺ = 2 x 10^-3 M
K_w = [H₃O⁺] [OH^–]

2 x 10^-3 >> 0.5 x 10^-11
i.e., [H₃O⁺] >> [OH^–], hence the juice is acidic in nature

Question 3.
Calculate the pH of 0.001M HCI solution
Answer:

H₃O from the auto ionisation of H₂O (10^-7 M) is negligible when compared to the H₃O from 10^-3 M HCI.
Hence [H₃O⁺] = 0.001 mol dm^-3
pH = – log₁₀ [H₃O⁺]
= – log₁₀ (0.001)
= – log₁₀ (10^-3) = 3

Question 4.
Calculate pH of 10^-7 M HCI
Answer:
If we do not consider [H₃O⁺] from the ionisation of H₂O, then [H₃O⁺] = [HCl] = 10⁷M i.e., pH =7, which is a pH of a neutral solution. We know that HCI solution is acidic whatever may be the concentration of HCI i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10^-7M). Hence, the H₃O⁺ (10^-7M) formed due to the auto ionisation of water cannot be neglected. So, in this case we should consider [H₃O⁺] from ionisation of H₂O
Answer:
[H₃O⁺] = 10^-7 (from HCl) + 10^-7(from water)
= 10^-7 (1+1)
= 2 x 10
pH = – log₁₀ [H₃O]
= – log₁₀(2 x 107) = – [log2 + log 10^-7]
= – log 2 – ( – 7). 1og₁₀
= 7 – log 2
= 7 – 0.3010 = 6.6990
= 6.70

Question 5.
A solution of 0.10 M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25°C. Find the dissociation constant of the acid.
Answer:
Given that α = 1.20% = \(\frac { 1.20 }{ 100 }\) = 1.2 x 10^-2
K_a = α²c = (1.2 X 10^-2)2 (0.1)
= 1.44 x 10^-4 x 10^-1 = 1.44 x 10^-5

Question 6.
Calculate the pH of 0.1M CH₃COOH solution. Dissociation constant of acetic acid is 1.8 x 10^-5.
Answer:
pH = – log[H⁺]
For weak acids,

Question 7.
Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. K_a for acetic acid is 1.8 x 10^-5

Given that K_a = 1.8 x 10^-5
pK_a = – log (1.8 x 10^-5)
= 5 – log 1.8
= 5 – 0.26 = 4.74
pH = 4.74 + log \(\frac { 0.20 }{ 0.18 }\)
= 4.74 + log \(\frac { 10 }{ 9 }\)
= 4.74 + log 10 – log 9
= 4.74 + 1 – 0.95
= 5.74 – 0.95
= 4.79

Question 8.
What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate and making the volume equal to 500 ml. (Given: K for acetic acid is 8 x 10)
Answer:
According to Henderson – Hessalbalch equation,

Given that K_a = 1.8 x 10^-5

pH = 4.74 + log 1
pH = 4.74 + 0 = 4.74

Question 9.
Calculate

• the hydrolysis constant,
• degree of hydrolysis and
• pH of O.1 M CH3COONa solution (pK_a for CH₃COOH is 4.74).

Answer:
CH₃COONa is a salt of weak acid (CH₃COOH) and a strong base (NaOH). Hence, the solutions is alkaline due to hydrolysis.
CH₃COO (aq) + H₂O (aq) \(\rightleftharpoons\) CH₃COOH (aq) + OH^– (aq)
Give that pK_a = 4.74
pK_a = – log K_a
i.e., K_a = antilog of ( – PK_a)
= antilog of ( – 4.74)
= antilog of( – 5 + 0.26)
10^-5 x 1.8

Question 10.
Establish a relationship between the solubility product and molar solubility for the following

1. BaSO₄
2. Ag₂(CrO₄)

Answer:

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Multiple Additional Questions

Samacheer Kalvi 12th Chemistry Ionic Equilibrium 1 Marks Questions and Answers

I. Choose the best answer.

Question 1.
Which one of the following buffer is present in blood?
(a) CH₃COOH + CH₃COONa
(b) NH₄OH + NH₄Cl
(c) H₂CO₃ + NaHCO₃
(d) HCI + NaCl
Answer:
(c) H₂CO₃ + NaHCO₃

Question 2.
Which of the following is mostly used in fertilizer industry?
(a) Lactic acid
(b) Sulphuric acid
(c) Tannic acid
(d) Carbonic acid
Answer:
(b) Sulphuric acid

Question 3.
Which of the following is present in an antacid tablet?
(a) NaOH
(b) Mg(OH)₂
(c) Al(OH)₃
(d) either (b) or (c)
Answer:
(d) either (b) or (c)

Question 4.
The acid present in milk is …………..
(a) Lactic acid
(b) Tannic acid
(c) Tartaric acid
(a) Acetic acid
Answer:
(a) Lactic acid

Question 5.
Consider the following statements.
(i) Acid tastes sour
(ii) Acid turns red litmus to blue
(iii) Acid reacts with metals and liberates hydrogen gas

Which of the above statement is I are correct?
(a) (i) only
(b) (i) & (ii)
(c) (i) & (iii)
(d) (ii) only
Answer:
(c) (i) & (iii)

Question 6.
Consider the following statements.
(i) Acid tastes sour.
(ii) Acid turns blue litmus to red
(iii) Acid has a tendency to accept a proton from other substances.

Which of the above statement is I are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) only
(a) (ii) only
Answer:
(c) (iii) only

Question 7.
Which of the following can act as an acid as well as base by Lowry – Bronsted theory?
(a) H₂O
(b) NH₃
(c) NH₄OH
(d) Ca(OH)₂
Answer:
(a) H₂O

Question 8.
In the reaction HCI + H₂O \(\rightleftharpoons\) H₃O + Cl^– which one of the acid-base pair?
(a) HCl + H₃O⁺
(b) HCI + CI^–
(c) H₃O + Cl
(d) H₂O + Cl^–
Answer:
(b) HCI + CI^–

Question 9.
Which of the following is considered as Lewis acid?
(a) NH₃
(b) BF₃
(c) HF
(d) HCl
Answer:
(b) BF₃

Question 10.
Which of the following is considered as Lewis base?
(a) BF₃
(b) AICI₃
(c) HCI
(d) NH₃
Answer:
(d) NH₃

Question 11.
Consider the following statements.
(i) A Lewis acid is a species that accepts an electron pair.
(ii) A Lewis acid is a species that donates an electron pair.
(iii) The ligand act as Lewis base.
Which of the above statement is / are not correct?
(a) (i) only
(b) (ii) only
(c) (I) & (ii)
(d) (ii) & (iii)
Answer:
(b) (ii) only

Question 12.
In [Cr(H₂O)₆]³⁺ which one of the following acts as Lewis acid?
(a) Cr
(b) Cr³⁺
(c) (HO)₆
(d) Cr^3-
Answer:
(b) Cr³⁺

Question 13.
In [Cr(H₂O)₆]³⁺ which one of the following acts as Lewis base?
(a) H₂O
(b) H₃O⁺
(c) Cr³⁺
(d) Cr
Answer:
(a) H₂O

Question 14.
Among the following which is the strongest acid?
(a) Formic acid
(b) Acetic acid
(c) Hydrochloric acid
(d) Lactic acid
Answer:
(c) Hydrochloric acid

Question 15.
Which of the following is the weak acid?
(a) HCl
(b) H₂SO₄
(c) HNO₃
(d) CH₃COOH
Answer:
(d) CH₃COOH

Question 16.
Identify the weakest acid?
(a) H₃O⁺
(b) H₂ SO₄
(c) OH^–
(d) CH₃COOH
Answer:
(c) OH^–

Question 17.
Which one of the following is the very weak base?
(a) NO₂^–
(b) NO₃^–
(c) NH₂^–
(d) O^2-
Answer:
(b) NO₃^–

Question 18.
Which one of the following is the strong base?
(a) ClO₄^–
(b) HSO₄
(c) O^2-
(d) F^–
Answer:
(c) O^2-

Question 19.
Which of the following is the weak base?
(a) H^–
(b) OH^–
(c) HSO₄
(d) F^–
Answer:
(d) F^–

Question 20.
The value of ionic product of water at 25°C is ………………..
(a) 1 x 10^-7
(b) 1 x 10⁷
(c) 1 x 10^-14
(d) 1 x 10¹⁴
Answer:
(c) 1 x 10^-14

Question 21.
Consider the following statements.
(i) The dissociation of water is an exothermic reaction.
(ii) With the increase in temperature, the ionic product of water value decreases.
(iii) With the increase in temperature, the ionic product of water value increases.
Which of the above statement is / are correct?
(a) (i) and (ii)
(b) (ii) only
(c) (iii) only
(d) (ii) & (iii)

Question 22.
Which of the following is a neutral solution?
(a) Aqueous NaCl solution
(b) Aqueous NaOH solution
(c) Aqueous HCl solution
(d) Aqueous NH₃
Answer:
(a) Aqueous NaCl solution

Question 23.
The pH of a neutral solution is ………..
(a) less than 7
(b) more than 7
(c) equal to 7
(d) 14
Answer:
(a) less than 7

Question 24.
If the pH of a solution is less than 7, it is called ………………… solution.
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Answer:
(b) Acidic

Question 25.
If the pH of a solution is more than 7, it is called ……………. solution.
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Answer:
(a) Basic

Question 26.
The pH value of water is …………..
(a) 14
(b) 7
(c) 3
(d) 1
Answer:
(b) 7

Question 27.
The pH of drain cleaner is ………..
(a) 7
(b) 1
(c) 14
(d) 0
Answer:
(c) 14

Question 28.
The pH of the battery acid is ………….
(a) 7
(b) 1
(c) 14
(d) 0
Answer:
(d) 0

Question 29.
The pH of 0.001 M HCI solution is ………
(a) 3
(b) 2
(c) 1
(d) 11
Answer:
(a) 3
Solution:
[H₃O⁺] = 0.001 mol dm^-3
pH = – log₁₀ [H₃O⁺] = – log₁₀ [0.001]
= – 1og₁₀ [10^-3] = 3
pH = 3

Question 30.
The pH of 0.01 M HCl solution is …………..
(a) 3
(b) 2
(c) 1
(d) 10
Answer:
(b) 2
Solution:
[H₃O⁺] = 0.01 M
pH = – 1og₁₀ [H₃O⁺] = – log₁₀ [0.01]
= – 1og₁₀[10^-3] = 3
pH = 3

Question 31.
What is the pH of 0.1 M HCI solution?
(a) 1
(b) 2
(c) 13
(d) 3
Answer:
(a) 1
Solution:
[H₃O⁺] = 0.1 M
pH = – log₁₀[H₃O⁺] = – log₁₀[0.1]
= – log₁₀[10^-1]
pH = l

Question 32.
Consider the following statements.
(i) Degree of dissociation (a) is the fraction of the total number of moles of a substance that dissociates at equilibrium.
(ii) When the dilution increases by 100 times, the dissociation increases by 100 times.
(iii) When the dilution increases by 100 times, the dissociation increases by 10 times.
Which of the above statement is I are correct?
(a) (ii) only
(b) (i) & (iii)
(c) (iii) only
(d) (I) only
Answer:
(a) (ii) only

Question 33.
Which of the following is not a buffer solution?
(a) CH₃COOH + CH₃COONa
(b) NH₄OH + NH₄Cl
(c) H₂CO₃ + NaHCO₃
(d) NaOH + NaCI
Answer:
(d) NaOH + NaCI

Question 34.
The mathematical expression of buffer capacity is …………

Answer:

Question 35.
Which one of the following is not correct?
(a) pH+pOH = 14
(b) pH = 7 +\(\frac { 1 }{ 2 }\) pK_a – \(\frac { 1 }{ 2 }\) pK_b
(c) pH x pOH = 1 x 10¹⁴
(d) pH = – log₁₀ [H₃O⁺]
Answer:
(c) pH x pOH = 1 x 10¹⁴

Question 36.
The chemical present in kidney as stones is …………..
(a) CaCl₂
(b) Ca(CO₃)₂
(c) Calcium nitrate
(d) Calcium oxalate
Answer:
(d) Calcium oxalate

Question 37.
If the pH of an aqueous solution is 7, the solution is …………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(c) neutral

Question 38.
Cl^– is the conjugate base of …………..
(a) HClO₄
(b) HCI
(c) ClO₄^–
(d) HClO₃
Answer:
(b) HCI

Question 39.
The conjugate base of H₂O and H₂SO₄ are …………..
(a) OH^– and HSO₄
(b) H₄O and SO₄^2-
(c) OH and SO₄^2-
(d) H₃O and HSO₄
Answer:
(a) OH^– and HSO₄

Question 40.
The K_sp of AgI is 1.5 x 10^-16. On mixing equal volume of the following solutions, precipitation will occur only with ………….
(a) 10^-7MAg⁺ and 10^-19M I^–
(b) 10^-8MAg⁺ and 10^-8M I^–
(c) 10^-16 M Ag⁺ and 10^-16 M I^–
(d) 10^-9 M Ag⁺ and 10^-9 M I^–
Answer:
(b) 10^-8MAg⁺ and 10^-8M I^–
K_sp of AgI = 1.5 x 10^-16
10^-8 M Ag⁺ and 10^-8 M I
Ionic product = 10^-16 = K_sp

Question 41.
The strongest Bronsted base in the following anion is ……………
(a) ClO^–
(b) ClO^2-
(c) ClO^3-
(d) ClO^4-
Answer:
(a) ClO^–
Solution:
HClO is the weakest acid and its conjugate base ClO^– is the strongest base.

Question 42.
Calculate the hydrolysis constant of the salt containing NO₂.
Given the K_a for HNO₂ = 4.5 x 10^-10
(a) 2.22 x 10^-5
(b) 2.02 x 10⁵
(c) 4.33 x 10⁴
(d) 3.03 x 10^-5
Answer:
(a) 2.22 x 10^-5
Solution:

Question 43.
Electrophiles are usually ……….
(a) Lewis acid
(b) Lewis base
(c) Bronsted acid
(d) Bronted base
Answer:
(a) Lewis acid
Solution:
Lewis acid are electrophile because they accept electron pair.

Question 44.
Which one is a Lewis acid?
(a) CIF₃
(b) H₂O
(c) NH₃
(d) OH
Answer:
(a) CIF₃
Solution:
CIF₃ have vacant d-orbital in central atom.

Question 45.
An aqueous solution of ammonium acetate is …………….
(a) faintly acidic
(b) faintly basic
(c) fairly acidic
(d) Almost neutral
Answer:
(d) Almost neutral
Solution:
It is a salt of weak acid and weak base.

Question 46.
The dissociation constant of a weak acid is 1.0 x 10^-10. The equilibrium constant for the reaction with strong base is …………
(a) 1.0 x 10^-5
(b) 1.0 x 10^-9
(c) 1.0 x 10⁹
(d) 1.0 x 10¹⁴
Answer:
(c) 1.0 x 10⁹
Solution:

Question 47.
Arrange the acids
(i) H₂SO₃
(ii) H₃PO₃ and
(iii) HClO₃in the decreasing order of acidity.
(a) (i) > (iii) > (ii)
(b) (i) > (ii) > (iii)
(c) (ii) > (iii) > (I)
(d) (iii) > (i)> (ii)
Answer:
(d) (iii) > (i)> (ii)
Solution:
Acidity is directly proportional to oxidation number. As the oxidation number of S, P and Cl in H₂SO₃, H₃PO₃ and HCIO₃ is +4, +3, +5 respectively. So decreasing order of acidity will
be (iii) > (I) > (ii)

Question 48.
The pH of 0.1 M solution of a weak monoprotic acid 1% ionised is ………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Solution:
Conc = 0.1 M
α = 1
= 0.1 x \(\frac { 1 }{ 100 }\) = 10^-3
[H⁺] = 10^-3
∴ pH = 3

Question 49.
K_sp for Cr(OH)₃ is 2.7 x 10^-3. What is the solubility in moles / litre?
(a) 1 x 10^-8
(b) 8 x 10^-8
(c) 1.1 x 10^-8
(d) 0.18 x 10^-8
Answer:
(b) 8 x 10^-8
Solution:

Question 50.
pK_a of acetic acid is 4.74. The concenttation of CH₃COONa is 0.01 M. The pH of CH₃COONa
is …………..
(a) 3.37
(b) 4.37
(c) 4.74
(d) 0.474
Answer:
(a) 3.37
Solultion:
[H⁺] = c.α = \(\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
pH = – log \(\left(\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}\right)^{1 / 2}\)
= \(\frac { 1 }{ 2 }\) [-1ogK_a – log c]
= \(\frac { 1 }{ 2 }\) [4.74 – log 10^-2]
= \(\frac { 1 }{ 2 }\) [4.74 + 2] = 3.37
pH = 3.37

Question 51.
One litre of water contains 10 mol hydrogen ions. The degree of ionisation in water will be …………….
(a) 8 x 10^-7
(b) 0.8 x 10^-9
(c) 3.6 x 10^-7
(d) 3.6 x 10^-9
Answer:
(a) 8 x 10^-7
Solution:
1 litre of water contains 1000/18 mole.
So, degree of ionisation = \(\frac{10^{-7} \times 18}{1000}\) = 1.8 x 10^-7

Question 52.
If the solubility product of lead iodide (PbI₂) is 3.2 x 10^-8. Then its solubility in moles / litre will be ………….
(a) 2 x 10^-3.
(b) 4 x 10^-4
(c) 1.6 x 10^-5
(d) 1.8 x 10^-5
Solution:
K_sp = 4s³
4s³ = 3.2 x 10^-8
s = 2 x 10^-3M

Question 53.
The pH of a soft drink is 3.82. Its hydrogen ion concentration will be ……………
(a) 1.96 x 10^-2mol / L
(b) 1.96 x 1o^-3 mol / L
(c) 1.5 x 10^-4 mol / L
(d) 1.96 x 10^-1 mol / L
Answer:
(c) 1.5 x 10^-4 mol / L
Solution:
pH = 3.82 = – log₁₀[H⁺]
∴ [H⁺] = 1.5 x 10^-4 mol / litre

Question 54.
The pH of a solution at 25°C containing 0.10 M sodium acetate and 0.03 M acetic acid is …………..
(pK_a for CH₃COOH = 4.57)
(a) 4.09
(b) 5.09
(c) 6.10
(d) 7.09
Answer:
(b) 5.09
Solution:
pH = pK_a + log \(\frac { [salt] }{ [acid] }\)
= 4.57 + log \(\frac { 0.10 }{ 0.03 }\) = 5.09

Question 55.
A weak acid is 0.1% ionised in 0.1 M solution. Its pH is …………..
(a) 2
(b) 3
(c) 4
(d) 1
Answer:
(c) 4
Solution:
For a monobasic acid [H⁺] = c.α
= \(\frac { 1 }{ 2 }\) x 0.001 = 10^-4
pH = – log₁₀[10^-4] = 4

Question 56.
Which one of the following is not a buffer solution?
(a) 0.8 M H₂S + 0.8 M KHS.
(b) 2 M C₆H₅NH₂ + 2 M C₆H₅N
(c) 3 M H₂CO₃ + 3 M KHCO₃
(d) 0.05 M KCIO₄ + 0.05 M HCIO
Answer:
(d) 0.05 M KCIO₄ + 0.05 M HCIO
Hint. HClO₄ is a strong acid while buffer is a mixture of weak acid and its salt.

Question 57.
The pH of pure water or neutral solution at 50°C is …………… (pK_w = 13.2613 at 50°C)
(a) 7.0
(b) 7.13
(c) 6.0
(d) 6.63
Answer:
(d) 6.63
Solution:
[H⁺] [OH^–] = 10^-13.26
[H⁺] = [OH^–]
[H⁺] = \(\frac { { 10 }^{ \frac { -13.26 }{ 2 } } }{ 2 }\)
∴ pH = 6.63

Question 58.
Increasing order of acidic character would be ……………..
(a) CH₃COOH < H₂SO₄ < HCO₃
(b) CH₃COOH < H₂CO₃ < H₂SO
(c) H₂CO₃ < CH₃COOH < H₂SO₄
(d) H₂SO₄ < H₂CO₃ < CH₃COOH
Answer:
(d) H₂SO₄ < H₂CO₃ < CH₃COOH

Question 59.
What is the pH of 1 M CH₃COOH solution?. Ka of acetic acid is 1.8 x 10^-5. K = 10^-14 mol² litre².
(a) 9.4
(b) 4.8
(c) 3.6
(d) 2.4
Answer:
(a) 9.4
Solution:
CH₃COO + H₂O \(\rightleftharpoons\) CH₃COOH + OH^–
[OH^–] = c x h

= 2.35 x 10^-5
pOH = 4.62
pH + pOH = 14
pH = 14 – 4.62 = 9.38

Question 60.
4Na + O₂ → 2Na₂O
Na₂O + H₂O → 2NaOH
In the given reaction, the oxide of sodium is …………..
(a) Acidic
(b) Basic
(c) Amphoteric
(d) Neutral
Answer:
(b) Basic
Solution.
Na₂O form NaOH so that it is basic oxide.

Question 61.
The pH of 0.001 M NaOH will be ………….
(a) 3
(b) 2
(c) 11
(d) 12
Answer:
(c) 11
Solution:
0.001 M NaOH means [OH^–] 0.001 .
10^-3 pOH = 3
pH + pOH = 14
pH = 14 – 3 = 11

Question 62.
The addition of pure solid sodium carbonate to pure water causes …………….
(a) an increase in hydronium ion concentration
(b) an increase in alkalinity
(c) No change in acidity
(d) A decrease in hydroxide ion
Answer:
(b) an increase in alkalinity
Hint.
Adding Na₂CO₃ to water makes the solution basic and hence pH increases from 7.

Question 63.
When solid potassium cyanide is added in water then ……………
(a) pH will increase
(b) pH will decrease
(c) pH will remain the same
(d) electrical conductivity will not change
Answer:
(a) pH will increase
Hint:
KCN + H₂O \(\rightleftharpoons\) KOH + HCN.
KOH is a strong base and HCN is a weak acid. So pH will increase.

Question 64.
pH of a solution is 5. Its hydroxyl ion concentration is …………..
(a) 5
(b) 10
(c) 10^-5
(d) 10^-9
Answer:
(d) 10^-9
Solution:
pH = 5 means [H⁺] = 10^-5
pOH = 14 – pH = 14 – 5 = 9
[OH^–] = 10^-pOH = 10^-9

Question 65.
Which one of the following is a buffer?
(a) CH₃COOH + CH₃COONa
(b) CH₃COOH + CH₃COONH₄
(c) NaOH + NaCI
(d) CH₃COOH + NH₄CI
Answer:
(a) CH₃COOH + CH₃COONa

Question 66.
Which will have maximum pH?
(a) Distilled water
(b) 1 M NH₃
(c) 1 M NaOH
(d) Water saturated by chlorine
Answer:
(c) 1 M NaOH
Hint:
NaOH has maximum [OH^–] and minimum of [H⁺ ] and so maximum pH value.

Question 67.
pH of a solution is 9.5. The solution is …………..
(a) Neutral
(b) Acidic
(c) Basic
(d) Amphoteric
Answer:
(c) Basic
Solution:
If pH = 7 solution is neutral
pH < 7 solution is acidic
pH > 7 solution is basic

Question 68.
A solution has pH = 5, it is diluted 100 times, then it will become ……………….
(a) Neutral
(b) Basic
(c) unaffected
(d) more acidic
Answer:
(a) Neutral
Solution:
pH = 5 means [H⁺] = 10^-5
After dilution [H⁺] = 10^-5 / 100 = 10^-7 M
[H⁺] from H₂O cannot be neglected.
Total [H⁺] = 10^-7 + 10^-7 = 2 x 10^-7
pH = 7 – 0.3010 = 6.6990 = 7
pH = 7 (Neutral)

Question 69.
By adding a strong acid to the buffer solution, the pH of the buffer solution ……………..
(a) remains constant
(b) increases
(c) decreases
(d) becomes zero
Answer:
(a) remains constant

Question 70.
pH of human blood is 7.4. Then H⁺ concentration will be ……………..
(a) 4 x 10^-8
(b) 2 X 10^-8
(c) 4 x 10^-4
(d) 2 x 10^-4
Answer:
(a) 4 x 10^-8
pH = – log [H⁺]
7.4 = – log [H⁺]
7.4 = log 1 – log [H⁺]
log [H⁺] = log 1 – 7.4
log [H⁺] = 8.6
[H⁺] Antilog of 8.6
= 4 x 10^-8

Question 71.
The highest pH 14 is given by ……………….
(a) 0.1 M H₂SO₄
(b) 0.1 M NaOH
(c) 1 N NaOH
(d) 1 N HCl
Answer:
(a) 0.1 M H₂SO₄
Solution:
[OH^–] = 1
pOH = 0
pH + pOH = 14
pH = 14 – 0 = 14

Question 72.
Which of the following is not a Bronsted acid?
(a) CH₃NH₄
(b) CH₃COO^–
(c) H₂O
(d) HSO₄
Answer:
(b) CH₃COO^–
Hint:
Those substances which give a proton is called Bronsted acid, while CH₃COO^– doesn’t have a proton. So it is not a Bronsted acid.

Question 73.
Pure water is kept in a vessel and it remains exposed to atmospheric CO₂ which is absorbed, then its pH will be
(a) greater than 7
(b) less than 7
(c) equal to 7
(d) depends on ionic production of water
Hint:
CO₂ is acidic oxide which on dissolution in water develops acidic nature.

Question 74.
As the temperature increases, the pH of KOH solution ……………..
(a) will decrease
(b) will increase
(c) remains constant
(d) depends upon concentration of KOH solution
Answer:
(a) will decrease

Question 75.
The pH of millimolar HCl is ………….
(a) 1
(b) 3
(c) 2
(d) 4
Answer:
(b) 3
Solution:
pH = – log [H⁺]
[H⁺] = 10^-3
pH = log 1 – log [H⁺]
= log 1 – log 10^-3 = 3

Question 76.
The unit of ionic product of water K is ……………
(a) mol^-1 L^-1
(b) mol^-2 L^-2
(c) mol^-2 L^-1
(d) mol² L^-2
Answer:
(d) mol² L^-2

Question 77.
Review the equilibrium and choose the correct statement.
HClO₄ + H₂O \(\rightleftharpoons\) H₃O⁺ + ClO₄⁺
(a) HClO₄ is the conjugate acid of H₂O
(b) H₃O is the conjugate base of H₂O
(c) H₂O is the conjugate acid of H₃O
(d) ClO₄ is the conjugate base of HCIO₄
Answer:
(d) ClO₄ is the conjugate base of HCIO₄

Question 78.
Which of the following is the strongest conjugate base?
(a) CI^–
(b) CH₃COO
(c) SO₄^2-
(d) NO₂^–
Hint:
CH₃COO is a conjugate base of a weak acid.
CH₃COOH \(\rightleftharpoons\) CH₃COO + H⁺

Question 79.
Which one of the following substance has the highest proton affinity?
(a) H₂O
(b) H₂S
(c) NH₃
(d) PH₃
Answer:
(c) NH₃

Question 80.
Which of the following is the strongest Lewis acid?
(a) BI₃
(b) BBr₃
(c) BCI₃
(d) BF₃
Hint:
Larger the size of the halogen atom less is the back donation of electrons into empty 2p orbital of B.

Question 81.
Which of the following is the weakest acid?
(a) HF
(b) HCI
(c) HBr
(d) HI
Answer:
(a) HF
Hint:
HF does not give proton easily.

Question 82.
Among the following, the weakest Lewis base is …………
(a) H^–
(b) OH^–
(c) CI^–
(d) HClO₃^–
Hint:
CI^– is a conjugate base of strong acid HCI.

Question 83.
Which of the following is not a Lewis acid?
(a) BF₃
(b) AlCI₃
(c) HCl
(d) LiAIH₄
Hint:
It is a nucleophile and capable of donating electron pair and so it can act as Lewis base.

Question 84.
Which one of the following is called amphoteric solvent?
(a) Ammonium hydroxide
(b) Chloroform
(c) Benzene
(d) Water
Answer:
(d) Water
Hint:

Question 85.
Which of the following is non – electrolyte?
(a) NaCl
(b) CaCl₂
(c) C₁₂H₂₂O₁₁
(d) CH₃COOH
Hint:
C₁₂H₂₂O₁₁ is a sugar and non-electrolyte.

Question 86.
At infinite dilution, the percentage ionisation for both strong and weak electrolyte is ………….
(a) 1%
(b) 20%
(c) 50%
(d) 100%
Answer:
(d) 100%
Hint:
According to Ostwald’s dilution law, degree of ionisation is directly proportional to the dilution.

Question 87.
An acid HA ionises as HA \(\rightleftharpoons\) H⁺ + A^– The pH of 1.0 M solution is 5. its dissociation constant would be …………..
(a) 1 x 10^-5
(b) 1 x 10^-10
(c) 5
(d) 5 x 10⁸
Answer:
(b) 1 x 10^-10

Question 88.
Three reactions involving H₂PO₄ are given below.
(i) H₃PO₄ + H₂O → H₃O⁺ + H₂PO₄^–
(ii) H₂PO₄ + H₂O → HPO₄² + H₃O⁺
(iii) H₂PO₄ + OH^– → H₃PO₄ + O₂^–

In which of the above does H₂PO₄^– act as an acid.
(a) (i) only
(b) (ii) only
(c) (i) & (iii)
(d) (iii) only
Answer:
(b) (ii) only

Question 89.
Which of the following is not a Lewis acid?
(a) CO
(b) SiCl₄
(c) SO₃
(d) Zn²⁺
Answer:
(c) SO₃
CO does not contain vacant d-orbital.

Question 90.
A chemist dissolves an excess of BaSO₄ in pure water at 25°C if its K_sp = 1 x 10^-10 What is the concentration of Barium in the water?
(a) 10^-14 M
(b) 10^-5 M
(c) 10^-15 M
(d) 10^-6 M
Answer:
(d) 10^-6 M

Question 91.
On addition of ammonium chloride to a solution of ammonium hydroxide ……………
(a) dissociation of NH₄OH increases
(b) concentration of OH^– increases
(c) concentration of OH^– decreases
(d) concentration of NH₄ and OH^– increases
Hint:
Due to common ion effect.

Question 92.
The solubility product of a salt having a general formula MX₂ in water is 4 x 10^-2. The concentration of M²⁺ ions in the aqueous solution of the salt is ………………….
(a) 2.0 x 10^-6 M
(b) 1.0 X 10^-4 M
(c) 1.6 x 10^-4 M
(d) 4.0 x 10^-2 M
Answer:
(b) 1.0 X 10^-4 M
Solution:
MX₂ \(\rightleftharpoons\) M²⁺+ 2X^–
K_sp = (2s)² (s) = 4s³

= 1.0 x 10^-4M

Question 93.
The solubility of an aqueous solution of Mg(OH)₂ be x then its K_sp is ……………
(a) 4 x³
(b) 108 x⁵
(c) 27 x⁴
(d) 9 x
Answer:
(a) 4 x³

Question 94.
What is the correct representation of the solubility product constant of Ag₂CrO₄?
(a) [Ag⁺]² [CrO₄^-2]
(b) [Ag⁺] [CrO₄^-2]
(c) [2Ag⁺] [CrO₄^-2]
(d) [2Ag⁺]² [CrO₄^-2]
Answer:
(a) [Ag⁺]² [CrO₄^-2]
Ag₂CrO₄ \(\rightleftharpoons\) [2 Ag⁺] [CrO₄^-2]
Hence, K_sp = [Ag⁺]² + [CrO₄^-2]

Question 95.
What is the pH value of \(\frac { N }{ 100 }\) KOH solution?
(a) 10
(b) 3
(c) 2
(d) 11
Answer:
(d) 11
Solution.
10^-3 N KOH will give [OH^–] = 10^-3 M
pOH = 3
pH + pOH = 14
pH = 14 – 3 = 11

Question 96.
Which pair will show common ion effect?
(a) BaCI₂ + Ba(NO₃)₂
(b) NaCI + HCI
(c) NH₄OH + NH₄CI
(d) AgCN + KCN
Answer:
(c) NH₄OH + NH₄CI

Question 97.
The sotubility of AgCI will be minimum in ………….
(a) 0.00 1 M AgNO₃
(b) pure water
(c) 0.01 M CaCI₂
(d) 0.01 M NaCl
Answer:
(c) 0.01 M CaCI₂
Solution.
0.01 M CaCI₂ gives maximum CI^– ions to keep K_sp of AgCl constant, decrease in [Ag⁺] will be maximum.

Question 98.
Ionic product of water increases if ………….
(a) pressure is reduced
(b) H⁺ is added
(c) OH^– is added
(d) temperature increases
Answer:
(d) temperature increases
Solution:
K_w increases with increase ¡n temperature.

Question 99.
pH of water is 7. When a substance Y is added in water, the pH becomes 13. The substance Y is a salt of …………..
(a) strong acid and strong base
(b) weak acid and weak base
(c) strong acid and weak base
(d) weak acid and strong base
Answer:
(d) weak acid and strong base

Question 100.
Sodium chloride is purified by passing HCl gas in a impure solution of sodium chloride. It is based on ………………
(a) Buffer action
(b) Common ion effect
(c) Association of salt
(d) Hydrolysis of salt
Answer:
(b) Common ion effect

II. Fill in the blanks.

1. …………. theory does not explain the behaviour of acids and base in non aqueous solvents.
2. According to Lowry Bronsted theory, an acid is defined as a substance that has a tendency to …………. a proton and base is a substance that has a tendency to …………. a proton.
3. HCI and Cl are called …………. pairs.
4. A …………. acid is a positive ion or an electron deficient molecule.
5. …………. is an anion or neutral molecule that donates one lone pair of electrons.
6. The ligands act as …………. and the central metal atoms that accepts a pair of electrons behave as a ………….
7. Carbonium ion act as …………. and carbanion act as ………….
8. Acids with …………. greater than ten are considered as strong acids and less than one are called weak acids.
9. OH^– and H₂ are considered as ………….
10. ClO₄^–, Cl^–, HSO₄, NO₃^– are considered as ………….
11. …………. can act as an acid as well as base.
12. At 25°C, the value of K_w is equal to ………….
13. With the increase in temperature, K_w value is ………….
14. The dissociation of water is an …………. reaction.
15. Aqueous solution of HCl is …………. whereas aqueous solution of NH₃ is ………….
16. For neutral solutions, the concentration of [H₃O⁺] as well as [OH^–] is equal to at 25°C.
17. The pH of battery acid is equal to ………….
18. The pH of drain cleaner is equal to ………….
19. …………. is the fraction of the total number of moles of a substance that dissociates at equilibrium.
20. When the dilution increases by 100 times, the dissociation increases by ………….
21. When dilution…………., the degree of dissociation of weak electrolyte also increases.
22. The buffer present in the blood is ………….
23. …………. introduced a quantity called buffer index ? as a quantitative measure of the
24. When an acid reacts with a base, a salt and water are formed and the reaction is called ………….
25. …………. is the conjugate base of the weak acid CH₃COOH.
26. Kidney stones are developed over a period of time due to the precipitation of ………….
27. The pH of sea water is …………. than 7.
28.  O^2- and H^– are ………….
29. All metal ions (or) atoms are ………….
30. All anions are ………….

Answers:

1. Arrhenius
2. donate, accept
3. Conjugate acid-base
4. Lewis
5. Lewis base
6. Lewis base, Lewis acid
7. Lewis acid, Lewis base
8. Ka value
9. very weak acids
10. very weak base
11. Water
12. 1 x 10^-14
13. increases
14. endothermic
15. acidic, basic
16. 1 x 10^-17
17. zero
18. 14
19. degree of dissociation ?
20. 10 times
21. increases
22. H₂CO₃ and NaHCO₃
23. Vanslyke, buffer capacity
24. Neutralization
25. CH₃COO^–
26. Calcium oxalate
27. greater
28. strong base
29. Lewis acids
30. Lewis bases

III. Match the following column – I & II using the correct code given below that.

Question 1.

Answer:
(a) 4 3 1 2

Question 2.

Answer:
(b) 3 1 4 2

Question 3.

Answer:
(c) 2 3 4 1

Question 4.

Answer:
(d) 4 3 2 1

Question 5.

Answer:
(a) 3 1 4 2

Question 6.

Answer:
(a) 2 3 4 1

Question 7.

Answer:
(b) 2 4 1 3

Question 8.

Answer:
(c) 3 1 4 2

Question 9.

Answer:
(d) 3 4 1 2

Question 10.

Answer:
(a) 2 1 4 3

Question 11.

Answer:
(a) 3 4 2 1

IV. Assertion and reasons.

Question 1.
Assertion(A): In the process of dissolution of HCl in water, HCl act as acid and H₂O act as base.
Reason (R): When HCl is dissolved in water, it donates a proton to water.
(a) Both A and R are correct and R explains A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A

Question 2.
Assertion(A): When ammonia dissolved in water, H20 acts as an acid.
Reason (R): When ammonia is dissolved in water, it accepts a proton from water. According to Lowry – Bronsted theory, proton donor is acid and so water act as an acid.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct. .
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 3.
Assertion (A): In the reaction HCl + H₂O \(\rightleftharpoons\) H₃O⁺ + Cl^–, HCl and Cl^– are conjugate acid – base pair.
Reason (R): By Lowry – Bronsted theory, chemical species that differ only by a proton are called conjugate acid – base pair.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 4.
Assertion(A): BF₃ is a Lewis acid.
Reason (R): Boron has a vacant 2p orbital to accept the lone pair of electrons donated by any substance to form a new coordinate covalent bond.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong .
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 5.
Assertion(A): in coordination compounds, the ligands acts as Lewis acid aid the central metal atom or ion act as Lewis base.
Reason (R): Ligands are capable of accepting of a pair of electrons donated by the central metal atom or ion.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are wrong

Question 6.
Assertion(A): SiF₄ can act as Lewis acid.
Reason (R): In SiF₄, the central atom can expand its octet due to the availability of empty d – orbitais and can accept a pair of electrons.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 7.
Assertion(A): NH₃, H₂O, ROH all are examples of Lewis bases.
Reason (R): Molecules with one or more lone pairs of electrons act as Lewis bases.
(a) Both A and R are wrong
(b) A is correct and R is the correct explanation of A.
(c) A is wrong but R is correct.
(d) A is correct but R is wrong
Answer:
(b) A is correct and R is the correct explanation of A.

Question 8.
Assertlon(A): HCl is an strong acid while HCOOH is a weak acid.
Reason (R): HCI is completely ionised in water whereas HCOOH is paritally ionised in water.
(a) Both A and R are wrong
(b) A is wrong but R is correct
(c) A is correct but R is wrong
(d) Both A and R are correct and R is the correct explanation of A
Answer:
(d) Both A and R are correct and R is the correct explanation of A

Question 9.
Assertion(A): With the increase in temperature, the ionic product of water also increases.
Reason (R): The dissociation of water is an endothermic reaction.
(a) Both A and R are correct and R is the correct explanation of A
(b) A is correct but R is wrong
(c) A is wrong but R is correct
(d) Both A and R are wrong
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 10.
Assertion(A): When dilution increases, the degree of dissociation of weak electrolyte also increases.
Reason (R): The degree of dissociation a is inversely proportional to concentration c. When the dilution increases by 100 times, the dissociation increases by 10 times.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 11.
Assertion(A): The addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid.
Reason (R): This is due to common ion effect. i.e., CH₃COOH and CH₃COONa both contains CH₃COO^– ion as common.
(a) Both A and R are correct and R is the correct explanation of A
(b) BothA and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 12.
Assertion(A): The solution of NH₄CI has pH value less than 7.
Reason (R): The salt of weak base (NH₄OH) and strong acid (HCl) is acidic in nature, when dissolved in water. So pH value is less than 7.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 13.
Assertion(A): pH = 7 signifies pure water.
Reason (R): pH = 7 means it is a neutral solution where [H₃O⁺] [OH^–]
(a) A is correct but R is wrong
(b) A is wrong but R is correct
(c) Both A and R are wrong
(d) A is correct and R does not explain A
Answer:
(b) A is wrong but R is correct

Question 14.
Assertion(A): A mixture of CH₃COOH and CH₃COONH₄ is an acidic buffer.
Reason (R): An acidic buffer contains a weak acid and the salt of weak acid with strong base.
(a) A is correct but R is wrong.
(b) A is wrong but R is correct.
(c) Both A and R are correct and R is the correct explanation of A
(d) Both A and R are wrong
Answer:
(b) A is wrong but R is correct.

Question 15.
Assertion(A): Buffer mixture is the one whose pH remains constant even by addition of strong acid or strong base.
Reason (R): To resist changes in its pH on the addition of an acid or base, the buffer solution should contain both acidic as well as basic components so as to neutralise the effect of added acid or base.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

V. Find the odd one out and give the reasons.

Question 1.
(a) HNO₃
(b) Ba(OH)₂
(c) H₃PO₄
(d) CH₃COOH
Answer:
(b) Ba(OH)₂
Reason: Ba(OH)₂ is the base whereas the others are acids.

Question 2.
(a) NH₃
(b) H₂O
(c) RNH₂
(d) BF₃
Answer:
(d) BF₃
Reason: BF₃ is a Lewis acid whereas others are Lewis base.

Question 3.
(a) SiF₄
(b) SF₄
(c) FeCl₃
(d) NH₃
Answer:
(d) NH₃
Reason: NH₃ is a Lewis base whereas others are Lewis acid.

Question 4.
(a) HCl
(b) H₂SO₄
(c) CH₃COOH
(d) HNO₃
Answer:
(c) CH₃COOH
Reason: CH₃COOH is a weak acid whereas others are strong acids.

Question 5.
(a) HCOOH
(b) CH₃COOH
(c) Lactic acid
(d) HCI
Answer:
(d)HCI
Reason: HCl is a strong acid whereas others are weak acids.

Question 6.
(a) HClO₄
(b) HCI
(c) HSO₄
(d) H₂SO₄
Answer:
(c) HSO₄.
Reason: HSO₄ is a very weak base whereas others are strong acid.

Question 7.
(a) NH₂^–
(b) O₂^–
(c) H^–
(d) OH^–
Answer:
(d) OH^–
Reason: OH^– is a very weak acid whereas others are strong bases.

Question 8.
(a) HNO₂
(b) HF
(c) H₂SO₄
(d) CH₃COOH
Answer:
(c) H₂SO₄
Reason: H₂SO₄ is a strong acid whereas others are weak acids.

Question 9.
(a) F^–
(b) CH₃COO
(c) O₂
(d) NO₂^–
Answer:
(c) O^2-
Reason: O^2- is a strong base whereas others are weak bases.

Question 10.
(a) Vinegar
(b) Black coffee
(c) Sea water
(d) Orange juice
Answer:
(c) Sea water
Reason: Sea water is basic and has pH > 7 whereas others are acidic and have pH < 7.

Question 11.
(a) Baking soda
(b) Tomato
(c) Soapy water
(d) Drain cleaner
Answer:
(b) Tomato
Reason: Tomato has pH less than 7 and it is acidic whereas others have pH greater than 7 and they are basic.

Question 12.
(a) CH₃COOH + CH₃COONa
(b) NH₄OH + NH₄CI
(c) H₂CO₃ + NaHCO₃
(d) NaOH + NaCl
Answer:
(d) NaOH + NaCl
Reason: NaOH + NaCl is not a buffer mixture whereas others are buffer mixtures.

VI. Find out the incorrect pair.

Question 1.
(a) HNO₃, H₂SO₄
(b) Al(OH)₃ , Mg (OH)₂
(c) CH₃COOH, HCOOH
(d) H₂O, OH^–
Answer:
(d) H₂O, OH

Question 2.
(a) HCl , Cl^–
(b) H₂O, H₃O⁺
(c) HNO₃ , HNO₂
(d) H₂SO₄, HSO₄^–
Answer:
(c) HNO₃ , HNO₂

Question 3.
(a) NH₃, H₂O
(b) ROH, ROR
(c) CN^–, SCN^–
(d) BF₃, H₂O
Answer:
(d) BF₃, H₂O

Question 4.
(a) BF₃ , BF₂
(b) Fe₂, Fe₃
(c) CaO, Mg(OH)₂
(d) SiF₄, SF₄
Answer:
(c) CaO, Mg(OH)₂

Question 5.
(a) Orange juice, Tomato juice
(b) Soapy water, Sea water
(c) Water, H₃O
(d) Bleach , Ammonia solution
Answer:
(c) Water, H₃O

VII. Find out the correct pair.

Question 1.
(a) HNO₃, Ba(OH)₂
(b) CH₃COOH , HCI
(c) H₃O⁺, Cl^–
(d) HCl + H₂SO₄
Answer:
(d) HCl + H₂SO₄.

Question 2.
(a) BF₃, NH₄⁺
(b) CH₂^– , CH₃⁺
(c) Fe²⁺, Fe³⁺
(d) (CH₃)3C⁺ , CH₂ = CH₂
Answer:
(c) Fe²⁺, Fe³⁺

Question 3.
(a) H₃O₊, HCI
(b) HSO₄, NO₂
(c) HNO₂, H₂
(d) HCl, Cl^–
Answer:
(d) HCI, CI^–

Question 4.
(a) Orange, Black coffee
(b) Baking soda,, Water
(c) Ammonia, Stomach acid
(d) Bleach , Tomato
Answer:
(a) Orange , Black coffee

Question 5.
(a) NH₄OH + NaOH
(b) NaOH + NaCl
(c) CH₃COOH + CH₃COONa
(d) CH₃COOH + CH₃COONH₄
Answer:
(c) CH₃COOH + CH₃COONa

VIII. Answer the following.

Question 1.
What are the general characteristics of acid and base?
Answer:

1. Acid tastes sour, turns the blue litmus to red and reacts with metals such as zinc and produces hydrogen gas.
2. Base tastes bitter, turns the red litmus to blue and soapy to touch.

Question 2.
Explain the Arrhenius concept of acid and base with example.
Answer:
1. According to Arrhenius concept, an acid is a substance that dissociates to give hydrogen ions in water. For example,

2. Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example,

Question 3.
What are the limitations of Arrhenius concept?
Answer:
1. Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.

2. This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

Question 4.
What is meant by strong acid and weak acid? Explain with example.
Answer:
1. A strong acid is the one that is almost completely dissociated in water.
HCl + H₂O → H₃O⁺ + Cl^–

2. A weak acid is the one that is partially dissociated in water.
CH₃COOH +H₂O \(\rightleftharpoons\) H₃O + CH₃COO^–

Question 5.
Give two examples for

1. Strong acid
2. Strong base

Answer:

1. Strong acid: HCIO₄, H₂SO₄
2. Strong base: NH₂^–, O^2-

Question 6.
Give two examples for

1. Very weak acid
2. Very weak base

Answer:

1. Very weak acid: OH^–,H₂
2. Very weak base: Cl^–, ClO₄^–

Question 7.
Give two examples for

1. Weak acid
2. Weak base

Answer:

1. Weak acid: HF, CH₃COOH
2. Weak base: F, CH₃COO

Question 8.
What is meant by auto ionisation of water?
Answer:
Pure water has a little tendency to dissociate. i:e., one water molecule donates a proton to another water molecule. This is known as auto ionisation of water.

Question 9.
Define – ionic product of water.
Answer:
H₂O + H₂O \(\rightleftharpoons\) H₃O + OH
The dissociation constant for the above ionisation is given as,

The concentration of pure water is one.
i.e., [H₂O]² = 1
K = [H₃O⁺] [OH^–]
K_w = ionic product of water.
K_w = 1 x 10^-14 at 25°C

Question 10.
K_w = 1 x 10^-14 at 25°C. Justify this statement.
Answer:
1. Experimentally found that the concentration of H₃O in pure water is 1 x 10^-7 at 25°C.

2. Since the dissociation of water produces equal number of H₃O₄ and OH^–, the concentration of OH^– is also equal to 1 x 10^– at 25°C. The ionic product of water at 25°C is
K_w = [H₃O⁺] [OH^–]
= [1 x 10^-7] [1 x 10^-7]
K_w = [1 x 10^-14]

Question 11.
With increase in temperature, K_w also increases. Why?
Answer:

1. All equilibrium constant K_w. is also a constant at a particular temperature.
2. The dissociation of water is an endothermic reaction.
3. With the increase in temperature, the concentration of H₃O⁺ and OH^– also increases and hence the ionic product also increases.

Question 12.
Aqueous HCl is an acidic solution whereas aqueous NH₃ is a basic solution. Justify this statement.
Answer:
HCI + H₂O \(\rightleftharpoons\) H₃O₊ + Cl^– in this case, in addition to auto ionisation of water, HCI molecule also produces H₃O ion by donating a proton to water and hence [H3OJ> [OH]. It means that the aqueous HCI solution is acidic. Similarly in basic solution such as aqueous NH₃, [OH^–] > [H₃O⁺] and it is basic.

Question 13.
What is the statement of Ostwald’s dilution law.
Answer:
When dilution increases, the degree of dissociation of weak electrolyte also increases.

This statement is known as Ostwald’s dilution law.

Question 14.
Define – Salt hydrolysis.
Answer:
Salts completely dissociate in aqueous solution to give their constituent ions. The ions so produced are hydrated in water. ¡n certain cases, the cation, anion ?r both react with water and the reaction is called salt hydrolysis. e.g.,
NaOH_(aq) + HNO_3(aq) → NaNO_3(aq) + H₂O₍₁₎

Question 15.
What is meant by conjugate acid-base pair? Find the conjugate acid / base for the following species
Answer:
HNO₂, CH^–, HCIO₄, OH^–, CO₃^2-, S^2-
An acid-base pair which differs by a proton only (HA \(\rightleftharpoons\) A^– + H⁺) is known as conjugate acid-base pair.
Conjugate acid: HCN, H₂O, HCO₃^–, HS^–.
Conjugate base: NO₂^–, ClO₄^–, O₂^–.

Question 16.
Which of the following are Lewis Acids?
H₂O, BF₃, H⁺ and NH₄⁺
Answer:
BF₃, H⁺ ions are Lewis acids.

Question 17.
What will be the conjugate bases for the Bronsted acids? HF, H₂SO₄ and H₂CO₃?
Answer:
Conjugate bases: F^–, HSO₄^–, HCO₃^–.

Question 18.
Write the conjugate acids for the following Bronsted bases:
NH₂^–, NH₃ and HCOO^–
Answer:
NH₃, NH₄ and HCOOH

Question 19.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Answer:

Question 20.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid I Lewis base?
(a) OH^– ions
(b) F^–
(c) H⁺
(d) BCI₃
Answer:
(a) OH^– ions can donate an electron pair and act as Lewis base.
(b) F^– ions can donate an electron pair and act as Lewis base.
(a) H⁺ ions can accept an electron pair and act as Lewis base.
(b) BCl₃ can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.

Question 21.
Predict the acidic, basic or neutral nature of the solutions of the following salts:
NaCI, KBr, NaCN, NH₄NO₃, NaNO₂, KF.
Answer:
NaCN, NaNO₂, KF solutions are basic, as they are salts of strong base, weak acid.
NaCl, KBr solutions are neutral, as they are salts, of strong acid, strong base.
NH₄NO₃ solution is acidic, as it is a salt of strong acid, weak base.

Question 22.
Ionic product of water at 310 K is 2.7 x 10^-14 What Is the pH of neutral water at this temperature?
Answer:

Question 23.
The aqueous solution of sugar does not conduct electricity whereas when sodium chloride is added to water, it conducts electricity. Justify this statement.
Answer:
1. Sugar is a non electrolyte and when it dissolves in water, there will be no ionisation takes place. If there is no free ions, it does not conduct electricity.

2. When sodium chloride is added to water, it is completely ionised to give Na ions and Cl^– ions. Due to the presence of ions, they will be possibility of electhcal conductance. Because ions are carriers of electric current.

Question 24.
A reaction betwen ammonia and boron trifluoride is given below

Identify the acid and base in the reaction. Which theory explain it?
Answer:

1. In the above reaction BF₃ is an acid and NH₃ is the base.

2. Lewis concept explain it as follows

3. A Lewis acid is an electron deficient molecule and capable of accepting a pair of electrons and a Lewis base is electron rich molecule and capable of donating a pair of electrons.

Question 25.
The salt of strong acid and strong base does not undergo hydrolysis. Explain.
Answer:
1. In this case, neither the cations nor the anions undergo hydrolysis. Therefore the solution remains neutral.

2. For example, in the aqueous solution of NaCl, its ions Na⁺ and Cl^– ions have no tendency to react with H⁺ or OH^– ions of water.

This is because the possible products of such reaction are NaOH and HCI which are completely dissociated. As a result, there is no change in the concentration of W and OH^– ions and hence the solution continues to remain neutral.

Samacheer Kalvi 12th Chemistry Ionic Equilibrium 3 Mark Questions ans Answers

Question 1.
Explain Lowry – Bronsted theory of acid and base.
Answer:
1. According to Lowry-Bronsted theory, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance.

2. An acid is a proton donor and a base is a proton acceptor.

3. When HCI is dissolved in H₂O, HCI donates a proton to H₂O. Thus HCI behaves as an acid and H₂O is a base.
HCI + H₂O \(\rightleftharpoons\) H₃O⁺ + Cl^–

Question 2.
Explain the reaction of water with ammonia by proton theory.
Answer:
1. When ammonia dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid.

2. The reaction is represented as

3. The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid base pairs Conjugate acid – base pair

Question 3.
Explain about the strength of acids on the basis of K_a value.
Answer:
1. K_a is called the ionisation constant or dissociation constant of the acid. It measures the strength of an acid.

2. Acids such as HCI, HNO₃ are almost completely? onised and hence they have high K_a value i.e., K_a for HCI at 25°C is 2 x 10⁶.

3. Acids such as formic acid and acetic acid are partially ionised in solution and have low K_a value. i.e., K_a for acetic acid 1.8 x 10^-5 at 25°C

4. Acids with K_a value greater than ten are considered as strong acids and less than one considered as weak acids.

Question 4.
Write 3 formulas of strong acids, strong bases and weak acids.
Answer:

1. HClO₄, HCI, H₂SO₄ – are strong acids
2. NH₂^–, O^2-, H^– – are strong bases
3. HNO₂, HF, CH₃COOH are weak acids

Question 5.
pH of a neutral solution is equal to 7. Prove it.
Answer:
1. in neutral solutions, the concentration of [H₃O⁺] as well as [OH^–] are equal to 1 x 10^-7M at 25°C.

2. The pH of a neutral solution can be calculated by substituting this [H₃O⁺] çoncentration in the expression
pH = – log₁₀ [H₃O⁺]
= – log₁₀ [1 x 10^-7]
= – ( – 7)log \(\frac { 1 }{ 2 }\) = + 7 (l) = 7

3. pH = 7 for a neutral solution

Question 6.
Derive the relation between pH and pOH
Answer:
pH = – log₁₀ [H₃O] ………………(1)
pOH = – log₁₀ [OH^–] ……………(2)
Adding equations (1) and (2),
pH + pOH = (- log₁₀ [H₃O⁺) + ( – log₁₀[OH^–])
= – [(log₁₀[H₃O]) + (log₁₀ [OH^–)]
pH + pOH = – log₁₀[H₃O⁺] [OH^–]
[H₃O] [OH^–] = K_w
pH + pOH = – log K_w
pH + pOH = pK_w
[pK_w = – 1og_10Kw]
At 25°C, the ionic product of water Kw = 1 x 10^-14.
pK_w = – 1og₁₀10^-14 = 14 log₁₀10 = 14
pK_w = 14
pH + pOH = 14 at 25°C.

Question 7.
When the dilution increases by 100 times, the dissociation increases by 10 times. Justify this statement.
Answer:
(i). Let us consideran acid with K_a value 4 x 10⁴. We are calculating the degree of dissociation of that acid at two different concentration 1 x 10^-2 M and 1 x 10^-4 M using Ostwalds dilution law

(iv) i.e., when the dilution increases by 100 times (concentration decreases from 1 x 10^-2M to 1 x 10^-4M), the dissociation increases by 10 times.

(v) When dilution increases, the degree of dissociation of weak electrolyte also increase. (Ostwalsd’s dilution law).

Question 8.
What Is buffer solution? Give an example for an acidic buffer and a basic buffer.
Answer:

1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.
2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH₄O and NH₄Cl.

Question 9.
Define buffer capacity and buffer index.
Answer:

1. The buffering ability of a solution can be measured in terms of buffer capacity.
2. Buffer index ?, as a quantitative measure of the buffer capacity.
3. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.
4. β = \(\frac { dB }{ d(pH) }\). dB = number of gram equivalents of acid / base added to one litre of buffer solution. d(pH) = The change in the pH after the addition of acid / base.

Question 10.
How is solubility product is used to decide the precipitation of ions?
Answer:
1. When the product of molar concentration of the constituent ions i.e., ionic product exceeds the solubility product then the compound gets precipitated.

2. When the
ionic Product > K_sp precipitation will occur and the solution is super saturated. ionic Product < K_sp no precipitation and the solution is unsaturated. ionic Product = K_sp equilibrium exist and the solution ¡s saturated.

3. By this way, the solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contain the constituent ions are mixed.

Question 11.
Derive the value of solubility product from molar solubility.
Answer:
1. Solubility can be calculated from molar solubility.i.e., the maximum number of moles of the solute that can be dissolved in one litre of the
solution.

2. For a solute X_mY_n
X_m Y_n(s) \(\rightleftharpoons\) mXⁿ⁺_(aq) + n Y^m-_(aq)

3. From the above stoichiometrically balanced equation, it is clear that I mole of X_m Y_n(s) dissociated to furnish ‘m’ moles of x and ‘n’ moles of Y. If’s’ is the molar solubility of X_m Y_nthen
Answer:
[Xⁿ⁺] = ms and [Y^m-] = ns
K_sp = [Xⁿ⁺]^m [Y^m-]ⁿ
K_sp = (ms)^m (ns)ⁿ
K_sp = (m)^m (n)ⁿ (s)^m+n

Question 12.
The concentration of hydrogen ions ¡n a sample of soft drink is 3.8 x 10^-3m. What is the. pH value? Whether the soft drink is acidic (or) basic?
Answer:
pH = – log₁₀ [H₃O⁺]
= – log₁₀ [3.8 x 10^–]
= – log 3.8 + 3
= 3 – 0.5798 = 2.4202
pH = 2.42
When pH < 7, the soft drink is acidic.

Question 13.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
pH = – log₁₀ [H₃O⁺]
= – log₁₀ = – pH = – 3.76
= \(\overline{4}\).24
[H₃O⁺] = antilog \(\overline{4}\).24
= l.738 x 10^-4
[H₃O⁺] = 1.74.x 10^-4M

Question 14.
The ionisation constant of HF, HCOOH, HCN at 298 K are 6.8 x 10^-4, 1.8 x 10^-4 and 4.8 x 10^-9 respectively. Calculate the ionisation constant of the corresponding conjugate base.
Answer:
1. HF, conjugate base is F
K_b = K_w/K_a = \(\frac{1 \times 10^{-4}}{6.8 \times 10^{-4}}\) = l.47 x 10^-11 = l.5 x 10^-11

2. for HCOO^–
K_b = \(\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}\) = 5.6 x 10^-11

3. for CN^–
K_b = \(\frac{1 \times 10^{-14}}{4.8 \times 10^{-4}}\) = 2.8 x 10 ^-6

Question 15.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
HCNO \(\rightleftharpoons\) H⁺ + CNO^–
pH = 2.34 means – log [H⁺] = 2.34 or log [H⁺] = – 2.34 = 3.86
or
[H⁺] = Antilog 3.86 = 4.57 x 10^-3 M
[CNO^–] = [H⁺] = 4.57 x 10^-3 M

Question 16.
The Ionization constant of nitrous acid is 4.5 x 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Sodium mtrite is a salt of weak acid, strong base. Hence,
K_h = 2.22 x 10_-11 K_w/K_b = 10^-14/(4.5x 10^-4)

[OH^–] = ch = 0.04 x 2.36 x 10^-5 = 944 x 10^-7
pOH = – log (9.44 x 10^-7) = 7 – 0.9750 = 6.03
pH = 14 – pOH = 14 – 6.03 = 7.97

Question 17.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K. For calcium sulphate, K_sp = 9.1 x 10^-6.
Answer:
CaSO₄(s) Ca²(aq) + SO^2-₄(aq)
If ‘s’ is the solubility of CaSO₄ in moles L^–, then K_sp = [Ca²⁺] x [SO₄^2-] = s²
or

= 3.02 x 10^-3 x 136gL^-1 = 0.411 gL^-1
(Molar mass of CaSO₄ = 136 g mol^-1)
Thus, for dissolving 0.441 g, water required = I L
For dissolving 1g, water required = \(\frac { 1 }{ 0.411 }\)L = 2.43L

Question 18.

1. Point out the differences between ionic product and solubility product.
2. The solubllity of AgCI in water at 298 K is 1.06 x 10^-5 mole per litre. Calculate is solubility product at this temperature.

Answer:
1. Ionic product

1. It is applicable to all types of solutions.
2. Its value changes with the change in con centration of the ions.

Solubility product

1. It is applicable to the saturated solutions.
2. It has a definite value for an electrolyte at a constant temperature.

2. The solubility equilibrium in the saturated solution is
AgCl (s) \(\rightleftharpoons\) Ag⁺(aq) + Cl^– (aq)
The solubility of AgCl is 1.06 x 10^-5 mole per litre.
[Ag⁺(aq)] = 1.06 x 10^-5 mol L^-1
[Cl^– (aq)] = 1.06 x 10^-5 mol L^-1
K_sp = [Ag⁺(aq)] [Cl^– (aq)]
= (1.06 x 10^-5 mol L^-1) x (1.06 x 10^-5 mol L^-1)
= 1.12 x 10^-2 moI² L^-2

Question 19.
The value of K of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 x 10^-15 and 6 x 10^-17 respectively. Which salt is more soluble? Explain.
Answer:

Ni(OH)₂ is more soluble than AgCN.

Question 20.
If 0.561 g KOH is dossolved in water to give 200 mL of solution at 298 K, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer:

Samacheer Kalvi 12th Chemistry Ionic Equilibrium 5 Marks Questions and Answers

Question 1.
Differentiate Lewis acids and Lewis bases.
Answer:
Lewis Acids

1. Lewis acids are substances that can accept one or more lone pair of electrons.
2. All metal ions (or) atoms can act as Lewis acids. Examples: Fe²⁺, Fe³⁺, Cu²⁺, Cr³⁺
3. Molecules that contain a polar double bond can act as Lewis acids. Examples: SO₂, CO₂, SO₃
4. Molecules in which the central atom can expand its act due to the availability of empty d-orbitais can act as Lewis acid. Example: SiF₄, SF₄, FeCI₃
5. Carbonium ion (CH₃)₃C⁺ can act as Lewis acid
6. Electron deficient molecules such as BF₃, AlCl₃, BeF₂ act as Lewis acid (electron pair acceptors)

Lewis Bases

1. Lewis bases are substances that can donate one or more lone pair of electrons.
2. All anions can act as Lewis bases. Examples: F^–, Cl^–, CN^–, SO₄^2-
3. Molecules that contain carbon-carbon multiple bond. Example: CH₂ = CH₂, CH = CH
4. All metal oxides can act as Lewis bases. Examples : CaO, MgO, Na₂O
5. CH₂^– carbanion cari act as Lewis acid
6. Electron rich molecules such as NH₃, H₂O, ROH, R – O – R, R – NH₂ act as Lewis base (Electron pair donors)

Question 2.
Explain about the ionisation of weak acid and how K₂ is derived?
Answer:
1. Weak acids are partially dissociated ¡n water and there is an equilibrium between the undissociated acid and its dissociated ions.

2. Consider the ionisation of weak monobasic acid HA in water
HA + H₂O \(\rightleftharpoons\) H₃O⁺ + A^– …………….(1)

3. Applying law of chemical equilibrium, the equilibrium constant K_c is given by the expression

4. In dilute solutions, water is present in large excess, hence its concentration may be taken as constant say K. Further H₃O⁺ indicates hydrated hydrogen ions, for simplicity, it may be replaced by H⁺. So the equation (2) becomes

5. The product of two constants K and K gives another constant. Let is be K₂

The constant K_a is called dissociation constant of weak acid.

Question 3.
Explain Buffer action with suitable example.
Answer:
Buffer action:
1. Let us consider buffer solution containing CH₃COOH and CH₃COO Na. The dissociation the buffer components occur as below.

2. If an acid is added to this mixture, it will be consumed by the conjugate base CH₃COO^– to form undissociated weak acid. i.e., the increase in the concentration of H⁺ does not reduce the pH significantly.
CH₃COO^–_(aq) + H⁺_(aq) → CH₃COOH_(aq)

3. If a base is added, it will be neutralised by H3O and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not altered.

4. On the addition of an acid (or) base to a buffer solution, there will be no change in its pH value. Because the buffer solution should contain both acidic as well as basic components so as to neutralise the effect of added acid (or) base at the same time, these components should not consume each other.

Question 4.
Prove the buffer action of acetic acid and sodium acetate by the addition of 0.01 mol of solid sodium hydroxide.
Answer:
1. Consider one litre of buffer solution containing 0.8 m CH₃COOH and O.8m CH₃COONa. Assume that the volume change due to the addition of 0.01 mol of solid NaOH is negligible. K_a for CH₃COOH is l.8 x 10^-5.

2.

3. The dissociation constant for CH₃COOH is given by

The above expression shows that the concentration of H⁺ is directly proportional to

degree of dissociation of CH₃COOH = α

4.

5. Given that K_a for CH₃COOH is 1.8 x 10^-5
[H⁺] = 1.8 x 10^-5
pH = – log [H⁺]
= – log [ 1.8 x 10^-5]
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

6. After adding 0.01 moI NaOH to I litre of buffer. Given that volume change due to the addition of NaOH is negligible. [OH^–] = 0.01 M. The consumption of OH^– are expressed by the following equation.

7. The addition of a strong base (0.01 M NaOH) increased the pH only slightly i.e., from 4.74 to 4.75. So the buffer action is verified.

Question 5.
DerIve Henderson – Hasselbalch equation
Answer:
1. The concentration of hydronium ion in acidic buffer solution depends on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution. i.e.,

2. The weak acid is dissociated only to a small extent. Moreover due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

3. [Acid] and [Salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution.

4. Taking logarithm on both sides of the equation

5. reverse the sign on both sides

Eqaution (6) & (7) are called Henderson – Hasselbalch equations.

Question 6.
Explain about the hydrolysis of salt of strong acid and a strong base with a suitable example.
Answer:
1. Let us consider the neutralisation reaction between NaOH and HNO₃ to give NaNO₃ and water.
NaOH_(aq) + HNO_3(aq) → NaNO_3(aq) + H₂O₍₁₎

2. The salt NaNO₃ completely dissociates in water to produce Na⁺ and NO₃ ions

3. Water dissociates to a small extent as
H₂O₍₁₎ H⁺_(aq) + OH^–_(aq)
Since [H⁺] = [OH^–], water is neutral.

4. NO₃ ion is the conjugate base of strong acid HNO₃ and hence it has no tendency to react withH⁺,

5. Similarly Na is the conjugate acid of the strong base NaOH and it has no tendency to react with OH

6. It means that there is no hydrolysis. In such cases [H⁺] (OH^–), pH is maintained and there fore the solution is neutral.

Question 7.
Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of K_h for that reaction.
Answer:
1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water.
NaOH_(aq) + CH₃COOH_(aq) \(\rightleftharpoons\) CH₃COONa_(aq) + H₂O₍₁₎

2. In aqueous solution, CH₃COONa is completely dissociated as follows.
CH₃COONa_(aq) CH₃COO^–_(aq) + Na⁺_(aq)

3. CH₃COO is a conjugate base of the weak acid CH₃COOH and it has a tendency to react with H⁺ from water to produce unionised acid. But there is no such tendency for Na⁺ to react with OH^–

4. CH₃COO^–_(aq) + H₂O(1) CH₃COOH_(aq) + OH^–₃ and therefore [OH^–] > [H⁺], in such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:

Equation (1) x (2)
K_h.K_a = [H⁺] [OH^–]
[H⁺] [OH^–] = K_w
K_h.K_a = K_w
K_h value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law K_h = h²C and [OH^–] =

Question 8.
DerIve the value of pH of salt solution in terms of K_a and concentration of electrolyte.
Answer:

Question 9.
Explain about the hydrolysis of salt of strong acid and weak base. Derive K_h and pH for that solution.
Answer:
1. Consider a reaction between strong acid HCl and a weak base NH₄OH to produce a salt NH₄CI and water

2. NH₄ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH- from water to produce unionised NH₄ as below,

3. There is no such tendency shown by Cl^– and therefore [H⁺] > [OH^–] the solution is acidic and the pH is less than 7.

4. In the salt hydrolysis of strong base and weak acid, we have to derive a relationship between K_h and K_b as
K_h . K_b = K_w

5.

Question 10.
Discuss about the hydrolysis of salt of weak acid and weak base and derive pH value for the solution.
Answer:
1. Consider the hydrolysis of ammonium acetate
CH₃COONH_4(aq) → CH₃COO^–_(aq) + NH⁺₄(aq)

2. In this case both the cation (NH₄⁺) and (CH₃COO^–) anion have the tendency to react with water.
CH₃COO^– + H₂O \(\rightleftharpoons\) CH₃COOH + OH^–
NH₄⁺ + H₂O \(\rightleftharpoons\) NH₄OH + H₃

3. The nature of the solution depends on the strength of acid (or) base i.e., if K_a > K_b, then the solution is acidic and pH < 7, if K_a < K_b then the solution is basic and pH > 7. If K_a = K_b, then the solution is neutral.

4. The relation between the dissociation constant K_a, K_b and hydrolysis constant is given by the following expression.
K_a . K_b. K_h = K_w

5. pH of the solution
pH = 7 + \(\frac { 1 }{ 2 }\)pK_a – \(\frac { 1 }{ 2 }\)pK_b

Question 11.
It has been found that the pH of a 0.01 M solution of an organic acid Is 4.15. Calculate the concentration of the anion, the Ionization constant of the acid and its pK_a.
Answer:
HA \(\rightleftharpoons\) H⁺
pH = log [H⁺] or log [H⁺] = – 4.15 = 5.85
[H⁺] = 7.08 x 10^-5 M = 7.08 x 10^-5 M
[A^–] = [H⁺] = 7.08 x 10^-5M

pK_a = – logK_a = – log (5.0 x 10^-7) = 7 – 0.699 = 6.301

Question 12.
Assuming complete dissociation, calculate the pH of the following solutions.
(i) 0.003 M HCl
(ii) 0.005 M NaOH
(iii) 0.002 M HBr
(iv) 0.002 M KOH
Answer:

Question 13.
What ¡s the pH of 0.001 M aniline solution? The ionisation constant of aniline is 4.27 x 10^-10. Calculate degree of ionization of aniline in the solution. Also calculate the ionisation constant of the conjugate acid of anile.
Answer:
1. C₆H₅NH₂ + H₂O \(\rightleftharpoons\) C₆H₅NH₃ + OH^–

2. Also

3. pK_a + pK_b = 14 (for a pair of conjugate acid and base)
pK_b = – log (4.27 x 10^-10) = 10 – 0.62 = 9.38
pK_a = 14 – 9.38 = 4.62
i.e., – log K_a 4.62 or log K_a = – 4.62 =

Question 14.
Calculate the degree of ionization of 0.05 M acetic acid If its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains

1. 0.01 M
2. 0.1 M HCI

Answer:
PK_a = i.e., – log K_a = 4.74
or log K_a = 4 . 74 = 5.26
K_a = 1.82 x 10^-5

In presence of HCI, due to high concentration of H⁺ ion, dissociation equilibrium will shift backward, Le., dissociation of acetic acid will decrease.
1. In presence of 0.01 M HCI, if x is the amount dissociated, then

2. In the presence of 0.1 M HCl, if y is the amount of acetic acid dissociated, then at equilibrium

Question 15.
The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.
Answer:

Common Errors

1. Acid and Base – Definiton
2. pH value
3. Buffer mixture
4. Conjugate Acid base pair

Rectifications

1. 
2. pH neutral
3. pH less than 7 – Acid
4. pH more than 7 – Base
5. Always either weak acid and its salt (or) weak base and its salt.
6. They differ by H⁺. For e.g., CH₃COOH. Its conjugate base is CH₃COO^–. H₂O – Acid and its conjugate base is OH^–

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Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Samacheer Kalvi 11th Accountancy Rectification of Errors Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Rectification Of Errors Questions With Answers Pdf Question 1.
Error of principle arises when ………………
(a) There is complete omission of a transaction
(b) There is partial omission of a transaction
(c) Distinction is not made between capital and revenue items
(d) There are wrong postings and wrong castings
Answer:
(c) Distinction is not made between capital and revenue items

Class 11 Accountancy Chapter 9 Solutions Question 2.
Errors not affecting the agreement of trial balance are ………………
(a) Errors of principle
(b) Errors of overcasting
(c) Errors of undercasting
(d) Errors of partial omission
Answer:
(a) Errors of principle

11th Accountancy Chapter 9 Book Back Answers Question 3.
The difference in trial balance is taken to ………………
(a) The capital account
(b) The trading account
(c) The suspense account
(d) The profit and loss account
Answer:
(c) The suspense account

Rectification Of Errors Questions With Solutions Class 11 Pdf Question 4.
A transaction not recorded at all is known as an error of ………………
(a) Principle
(b) Complete omission
(c) Partial omission
(d) Duplication
Answer:
(b) Complete omission

Rectification Of Errors Questions With Answers Question 5.
Wages paid for installation of machinery wrongly debited to wages account is an errs of ………………
(a) Partial omission
(b) Principle
(c) Complete omission
(d) Duplication
Answer:
(b) Principle

Accountancy Class 11 Chapter 9 Solutions Question 6.
Which of the following errors will not affect the trial balance?
(a) Wrong balancing of an account
(b) Posting an amount in the wrong account but on the correct side
(c) Wrong totalling of an account
(d) Carried forward wrong amount in a ledger account
Answer:
(b) Posting an amount in the wrong account but on the correct side

Rectification Of Errors Problems With Solutions Pdf Question 7.
Goods returned by Senguttuvan were taken into stock, but no entry was passed in the books. While rectifying this error, which of the following accounts should be debited?
(a) Senguttuvan account
(b) Sales returns account
(c) Returns outward account
(d) Purchases returns account
Answer:
(b) Sales returns account

Rectification Of Errors Questions With Solutions Class 11 Question 8.
A credit purchase of furniture from Athiyaman was debited to purchases account. Which of the following accounts should be debited while rectifying this error?
(a) Purchases account
(b) Athiyaman account
(c) Furniture account
(d) None of these
Answer:
(c) Furniture account

Rectification Of Errors In Tamil Question 9.
The total of purchases book was overcast. Which of the following accounts should be debited in the rectifying journal entry?
(a) Purchases account
(b) Suspense account
(c) Creditor account
(d) None of the above
Answer:
(b) Suspense account

Samacheer Kalvi 11th Accountancy Solutions Question 10.
Which of the following errors will be rectified using suspense account?
(a) Purchases returns book was undercast by ₹ 100
(b) Goods returned by Narendran was not recorded in the books
(c) Goods returned by Akila ₹ 900 was recorded in the sales returns book as ₹ 90
(d) A credit sale of goods to Ravivarman was not entered in the sales book
Answer:
(a) Purchases returns book was undercast by ₹ 100

II. Very Short Answer Questions

Samacheer Kalvi Guru 11th Accountancy Question 1.
What is meant by rectification of errors?
Answer:
Correction of errors in the books of accounts is not done by erasing, rewriting or striking the figures which are incorrect. Correcting the errors that has occured is called Rectification.

Chapter 9 Accountancy Class 11 Solutions Question 2.
What is meant by error of principle?
Answer:
It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.

Rectification Of Errors Meaning In Tamil Question 3.
What is meant by error of partial omission?
Answer:
When the accountant has failed to record a part of the transaction, it is known as error of partial omission. This error usually occurs in posting. This error affects only one account.

Class 11th Accounts Chapter 9 Solutions Question 4.
What is meant by error of complete omission?
Answer:
It means the failure to record a transaction in the journal or subsidiary book or failure to post both the aspects in ledger. This error affects two or more accounts.

Accountancy Chapter 9 Class 11 Question 5.
What are compensating errors?
Answer:
The errors that make up for each other or neutralize each other are known as compensating errors. These errors may occur in related or unrelated accounts. Thus, excess debit or credit in one account may be compensated by excess credit or debit in some other account. These are also known as offsetting errors.

III. Short Answer Questions

Rectification Of Errors Class 11 Solutions Question 1.
Write a note on error of principle by giving an example.
Answer:
It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.
Example:
Entering the purchase of an asset in the purchases book. Machinery purchased on credit for ₹ 10,000 by M/s. Anbarasi garments manufacturing company entered in the purchases book.

Question 2.
Write a note on suspense account.
Answer:
When the trial balance does not tally, the amount of difference is placed to the debit (when the total of the credit column is higher than the debit column) or credit (when the total of the debit column is higher than the credit column) to a temporary account is known as ‘suspense account’.

Question 3.
What are the errors not disclosed by a trial balance?
Answer:
Certain errors will not affect the agreement of trial balance. Though such errors occur in the books of accounts, the total of debit and credit balance will be the same. The trial balance will tally. Errors of complete omission, error of principle, compensating error, wrong entry in the subsidiary books are not disclosed by the trial balance.

Question 4.
What are the errors disclosed by a trial balance?
Answer:
Certain errors affect the agreement of trial balance. If such errors have occurred in the books of accounts, the total of debit and credit balances will not be the same. The trial balance will not tally. Error of partial omission and error of commission affect the agreement of trial balance.

Question 5.
Write a note on one – sided errors and two – sided errors.
Answer:

1. One – sided errors: When preparing the trial’balance, if the total of debit balances and credit balances are not the same, there is disagreement of trial balance.
2. Two – sided errors: Rectification of two – sided errors at the time of preparing the trial balance is just similar to that of their rectification before preparation of trial balance.

IV. Exercises

Question 1.
State the account/s affected in each of the following errors: (2 marks)
(a) Goods purchased on credit from Saranya for ₹ 150 was posted to the debit side of her account.
(b) The total of purchases book ₹ 4,500 was posted twice.
Answer:
(a) Purchases from Saranya should have been posted to the credit of Saranya’s A/c, but it has been debited. Hence, credit Saranya’s A/c with double the amount i.e., Rs. 300.
(b) Credit the Purchases A/c.

Question 2.
State the account/s affected in each of the following errors: (2 marks)
(a) Goods sold to Vasu on credit for ₹ 1,000 was not recorded in the sales book.
(b) The total of sales book ₹ 2,500 was posted twice.
Answer:
(a)

(b) Debit the Sales A/c

Question 3.
Rectify the following errors discovered before the preparation of the trial balance: (2 marks)
(a) Sales book was undercast by ₹ 100
(b) Purchases returns book was overcast by ₹ 200
Answer:
(a) Sales account should be credited with ₹ 100
(b) Purchases returns account should be debited with ₹ 200

Question 4.
Rectify the following errors before the preparation of trial balance: (3 marks)
(a) Returns outward book was undercast by ₹ 2,000
(b) Returns inward book total was taken as ₹ 15,000 instead of ₹ 14,000
(c) The total of the purchases account was carried forward ₹ 100 less.
Answer:
(a) Returns outward Account should be credited with ₹ 2,000
(b) Sales returns account should be credited with ₹ 1,000
(c) Purchases account should be debited with ₹ 100

Question 5.
Rectify the following errors assuming that the trial balance is yet to be prepared: (5 marks)
(a) Sales book was undercast by ₹ 400
(b) Sales returns book was overcast by ₹ 500
(c) Purchases book was undercast by ₹ 600
(d) Purchases returns book was overcast by ₹ 700
(e) Bills receivable book was undercast by ₹ 800
Answer:
(a) Sales account should be credited with ₹ 400
(b) Sales returns account should be credited with ₹ 500
(c) Purchases account should be debited with ₹ 600
(d) Purchases returns account should be debited with ₹ 700
(e) Bills receivable account should be debited with ₹ 800

Question 6.
Rectify the following errors before preparing trial balance: (5 marks)
(a) The total of purchases book was carried forward ₹ 90 less.
(b) The total of purchases book was carried forward ₹ 180 more
(c) The total of sales book was carried forward ₹ 270 less.
(d) The total of sales returns book was carried forward ₹ 360 more.
(e) The total of purchases returns book was carried forward ₹ 450 less.
Answer:
(a) Purchases account should be debited ₹ 90
(b) Purchases account should be credited ₹ 180
(c) Sales account should be credited ₹ 270
(d) Sales returns account should be credited with ₹ 360
(e) Purchases returns account should be credited with ₹ 450

Question 7.
The following errors were located by the accountant before preparation of trial balance. Rectify them. (5 marks)
(a) The total of the discount column of ₹ 1,100 on the debit side of the cash book was not yet posted.
(b) The total of the discount column on the credit side of the cash book was undercast by ₹ 500.
(c) Purchased goods from Anbuchelvan on credit for ₹ 700 was posted to the debit side of his account.
(d) Sale of goods to Ponmukil on credit for ₹ 78 was posted to her account as ₹ 87.
(e) The total of sales returns book of ₹ 550 was posted twice.
Answer:

Question 8.
The accountant of a firm located the following errors before preparing the trial balance. Rectify them. (5 marks)
(a) Machinery purchased for ₹ 3,000 was debited to purchases account.
(b) Interest received ₹ 200 was credited to commission account.
(c) An amount of ₹ 1,000 paid to Tamilselvan as salary was debited to his personal account.
(d) Old furniture sold for ₹ 300 was credited to sales account.
(e) Goods worth ₹ 800 purchased from Soundarapandian on credit was not recorded in the books of accounts.
Answer:
Rectifying Journal

Question 9.
Rectify the following errors which were located before preparing the trial balance. (5 marks)
(a) Wages paid ₹ 2,000 for the erection of machinery was debited to wages account.
(b) Sales returns book was short totalled by ₹ 1,000.
(c) Goods purchased for ₹ 200 was posted as ₹ 2,000 to purchases account.
(d) The sales book was overcast by ₹ 1,500.
(e) Cash paid to Mukil ₹ 2,800 which was debited to Akhil’s account as ? 2,000.
Answer:
Rectifying Journal

Question 10.
Rectify the following errors which were located at the time of preparing the trial balance: (5 marks)
(a) The total of the discount column on the debit side of the cash book of ₹ 225 was posted twice.
(b) Goods of the value of ₹ 75 returned by Ponnarasan was not posted to his account.
(c) Cash received from Yazhini ₹ 1,000 was not posted.
(d) Interest received ₹ 300 has not been posted.
(e) Rent paid ₹ 100 was posted to rent account as ₹ 10.
Answer:

Question 11.
The following errors were located at the time of preparing trial balance. Rectify them. (5 marks)
(a) A personal expense of the proprietor ₹ 200 was debited to travelling expenses account.
(b) Goods of ₹ 400 purchased from Ramesh on credit was wrongly credited to Ganesh’s account.
(c) An amount of ₹ 500 paid as salaries to Mathi was debited to his personal account.
(d) An amount of ₹ 2,700 paid for extension of the building was debited to repairs account.
(e) A credit sale of goods of ₹ 700 on credit to Mekala was posted to Krishnan’s account.
Answer:

Question 12.
Rectify the following journal entries. (5 marks)

Answer:

Question 13.
Rectify the following errors discovered after the preparation of the trial balance: (5 marks)
(a) Rent paid was carried forward to the next page ₹ 500 short.
(b) Wages paid was carried forward ₹ 250 excess.
Answer:
(a) Rent account is to be debited with ₹ 500.
(b) Wages account is to be credited with ₹ 250.

Question 14.
Rectify the following errors after preparation of trial balance: (5 marks)
(a) Salary paid to Ram ₹ 1,000 was wrongly debited to his personal account.
(b) A credit sale of goods to Balu for ₹ 450 was debited to Balan.
Answer:

Question 15.
Pass necessary journal entries to rectify the following errors located after the preparation of trial balance: (5 marks)
(a) Sales book was undercast by ₹ 1,000.
(b) An amount of ₹ 500 paid for wages was wrongly posted to machinery Account.
Answer:
(a) Sales account should be credited ₹ 1,000.
(b)

Question 16.
Give journal entries to rectify the following errors discovered after the preparation of trial balance: (5 marks)
(a) Purchases book was overcast by ₹ 10,000.
(b) Repairs to furniture of ₹ 500 was debited to furniture account.
(c) A credit sale of goods to Akilnilavan for ₹ 456 was credited to his account as ₹ 654.
Answer:
(a) Purchases account should be credited ₹ 10,000.

Question 17.
Rectify the following errors located after the preparation of trial balance: (5 marks)
(a) Purchases book was undercast by ₹ 900.
(b) Sale of old furniture for ₹ 1,000 was credited to sales account.
(c) Purchase of goods from Arul for ₹ 1,500 on credit was not recorded in the books.
Answer:
(a) Purchases account should be debited with ₹ 900.

Question 18.
The following errors were located after the preparation of trial balance. Pass journal entries to rectify them. Assume that there exists a suspense account. (5 marks)
(a) The total of sales book was undercast by ₹ 350.
(b) The total of the discount column on the debit side of cash book ₹ 420 was not posted.
(c) The total of one page of the purchases book of ₹ 5,353 was carried forward to the next page as ₹ 5,533.
(d) Salaries ₹ 2,400 was posted as ₹ 24,000.
(e) Purchase of goods from Sembiyanmadevi on credit for ₹ 180 was posted to her account as ₹ 1,800.
Answer:
Rectifying Journals

Question 19.
Rectify the following errors assuming that, the trial balance is already prepared and the difference was placed to suspense account: (5 marks)
(a) Sales book was undercast by ₹ 250.
(b) Purchases book was undercast by ₹ 120.
(c) Sales book was overcast by ₹ 130.
(d) Bills receivable book was undercast by ₹ 75.
(e) Purchases book was overcast by ₹ 35.
Answer:
(a) Sales account should be credited ₹ 250.
(b) Purchases account should be debited ₹ 120.
(c) Sales account should be debited ₹ 130.
(d) Bills receivable account should be debited ₹ 75.
(e) Purchases account should be credited ₹ 35.

Question 20.
The following errors were located after the preparation of trial balance. The difference in trial balance has been taken to suspense account. Rectify them. (5 marks)
(a) The total of purchases book was carried forward ₹ 70 less.
(b) The total of sales book was carried forward ₹ 340 more.
(c) The total of purchases book was carried forward ₹ 150 more.
(d) The total of sales book was carried forward ₹ 200 less.
(e) The total of purchases returns book was carried forward ₹ 350 less.
Answer:
Rectifying Journals

Question 21.
The following errors were located by the accountant after the preparation of trial balance. There exists a suspense account. Rectify them. (5 marks)
(a) The total of the discount column of ₹ 1,180 on the debit side of the cash book was not posted.
(b) Purchase of goods from Arivuchelvan on credit for ₹ 600 was posted to the debit side of his account.
(c) The total of the discount column on the credit side of the cash book was undercast by ₹ 400.
(d) The total of sales returns book of ₹ 570 was posted twice.
(e) Sold goods to Mukil on credit for ₹ 87 was posted to her account as ₹ 78.
Answer:

Question 22.
The accountant of a firm located the following errors after preparing the trial balance. Rectify them assuming that there is a suspense account. (5 marks)
(a) Machinery purchased for ₹ 3,500 was debited to purchases account.
(b) ₹ 1,800 paid to Raina as salary was debited to his personal account.
(c) Interest received ₹ 200 was credited to commission account.
(d) Goods worth ₹ 1,800 purchased from Amudhanila on credit was not recorded in the books of accounts.
(e) Used furniture sold for ₹ 350 was credited to sales account.
Answer:

Question 23.
The book – keeper of a firm found that the trial balance was out by ₹ 922 (excess credit). He placed the amount in the suspense account and subsequently found the following errors: (5 marks)
(a) The total of discount column on the credit side of the cash book ₹ 78 was not posted in the ledger.
(b) The total of purchases book was short by ₹ 1,000.
(c) A credit sale of goods to Natarajan for ₹ 375 was entered in the sales book as ₹ 735.
(d) A credit sale of goods to Mekala for ₹ 700 was entered in the purchases book. You are required to give rectification entries and prepare suspense account.
Answer:

Suspense Account

Question 24.
The books of Raman did not agree. The accountant placed the difference of ₹ 1,270 to the debit of suspense account. Rectify the following errors and prepare the suspense account:
(a) Goods taken by the proprietor for his personal use ₹ 75 was not entered in the books.
(b) A credit sale of goods to Shanmugam for ₹ 430 was credited to his account as ₹ 340.
(c) A purchase of goods on credit for ₹ 400 from Vivek was entered in the sales book. However, Vivek’s account was correctly credited.
(d) The total of the purchases returns book ₹ 300 was not posted.
Answer:

Suspense Account

Textbook Case Study Solved

Question 1.
Rameela, a class 11 student, visited one of her relative’s furniture shop. She met the accountant of the shop. He was busy with preparing final accounts. At that time, one of the staff approached the accountant with a list of errors found in ledger postings. Rameela asked the accountant, in a surprised tone, “Is it possible to rectify the errors before preparing the final accounts?” The accountant replied, “Yes, it is!” final accounts?” The accountant replied,
“Yes, it is!”
Rameela was curious to analyse the errors. She found the following:

1. Furniture sold on credit to Siva and company for ₹ 12,000 was debited to Sam and company;
2. Rent paid ₹ 2,500, was debited to rent account as ₹ 250.
3. The total of purchase journal was undercast by ₹ 1,000.
4. A sales invoice for ₹ 2,000, completely omitted from the books.
5. Stationery bought for ₹ 250, was posted to purchases account.

Can you help Rameela to identify and rectify the errors?
Answer:

Samacheer Kalvi 11th Accountancy Rectification of Errors Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
The errors can be classified into ……………… types.
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(d) Four

Question 2.
When the accountant has failed to record a part of the transaction is known as ………………
(a) Error of partial omission
(b) Error of commission
(c) Compensating errors
(d) Error of principle
Answer:
(a) Error of partial omission

Question 3.
The errors that make up for each other or neutralise each other are known as ………………
(a) Errors of commission
(b) Errors of principle
(c) Errors of omission
(d) Compensating errors
Answer:
(d) Compensating errors

Question 4.
The total of salary account is carried forward ₹ 1200 excess ……………….
(a) Errors in carry forward
(b) Errors in posting
(c) Errors in casting
(d) Errors of commission
Answer:
(a) Errors in carry forward

Question 5.
Sales book is undercast by ₹ 100, classify the error
(a) Errors of principle
(b) Errors of commission
(c) Errors in casting
(d) Errors of omission
Answer:
(c) Errors in casting

II. Very Short Answer Questions

Question 1.
What is error of omission?
Answer:
The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be complete omission or partial omission.

Question 2.
What is error of commission?
Answer:
When a transaction is incorrectly recorded, it is known as error of commission. It usually occurs due to lack of concentration or carelessness of the accountant.

Question 3.
What do you mean by errors?
Answer:
Errors means recording or classifying or summarising the accounting transactions wrongly or omissions to record them by a clerk or an accountant unintentionally.

III. Short Answer Questions

Question 1.
What are the types of errors at the stage of journalising?
Answer:

1. Error of omission
2. Error of commission
3. Error of principle

Question 2.
What are the types of errors at the stage of posting?
Answer:
(i) Errors of Omission:
(a) Error of complete omission
(b) Error of partial omission

(ii) Errors of Commission:
(a) Posting to wrong account
(b) Posting of wrong account
(c) Posting to the wrong side

Question 3.
What are the types of errors at the stage of preparing trial balance?
Answer:
(i) Error of Omission
(ii) Error of Commission:

(a) Entering to wrong account
(b) Entering wrong amount
(c) Entering to the wrong side of trial balance, etc.

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Tamil Nadu Samacheer Kalvi 10th Tamil Model Question Paper 2

நேரம்: 3.00 மணி
மதிப்பெண்கள்: 100

(குறிப்புகள்:

• இவ்வினாத்தாள் ஐந்து பகுதிகளைக் கொண்டது. அனைத்து பகுதிகளுக்கும் விடையளிக்க – வேண்டும். தேவையான இடங்களில் உள் தேர்வு வினாக்கள் கொடுக்கப்பட்டுள்ளது. காக
• பகுதி I, II, III, IV மற்றும் Vல் உள்ள அனைத்து வினாக்களுக்குத் தனித்தனியே விடையளிக்க வேண்டும்.
• வினா எண். 1 முதல் 15 வரை பகுதி-1ல் தேர்வு செய்யும் வினாக்கள் தரப்பட்டுள்ளன. ஒவ்வொரு வினாவிற்கும் ஒரு மதிப்பெண். சரியான விடையைத் தேர்ந்தெடுத்து குறியீட்டுடன் எழுதவும்.
• வினா எண் 16 முதல் 28 வரை பகுதி-IIல் இரண்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன: ஏதேனும் 9 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
• வினா எண் 29 முதல் 37 வரை பகுதி-IIIல் மூன்று மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. –
  ஏதேனும் 6 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
• வினா எண் 38 முதல் 42 வரை பகுதி-IVல் ஐந்து மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. ஏதேனும் 5 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
• வினா எண் 43 முதல் 45 வரை பகுதி-Vல் எட்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. அனைத்து வினாவிற்கும் விடையளிக்கவும்.

பகுதி – 1 (மதிப்பெண்கள்: 15)

(i) அனைத்து வினாக்களுக்கும் விடையளிக்கவும்.
(ii) கொடுக்கப்பட்ட நான்கு விடைகளில் சரியான விடையினைத் தேர்ந்தெடுத்துக் குறியீட்டுடன் விடையினையும் சேர்த்து எழுதுக. [15 x 1 = 15]

(குறிப்பு: விடைகள் தடித்த எழுத்தில் உள்ளன.)

Question 1.
மரஞ்செடியினின்று பூ கீழே விழுந்த நிலையைக் குறிக்கும் சொல் எது?
(அ) அரும்புஆ
(ஆ) மலர்
(இ) வீ
(ஈ) செம்மல்
Answer:
(இ) வீ

Question 2.
கூத்தராற்றுப்படை என்ற நூலில் ……………….. அடிகள் உள்ளன.
(அ) 383
(ஆ) 483
(இ) 583
(ஈ) 683
Answer:
(இ) 583

Question 3.
தொகைநிலைத் தொடர்……………. வகைப்படும்.
(அ ) மூன்று
ஆ) நான்கு
(இ) ஐந்து
(ஈ) ஆறு
Answer:
(ஈ) ஆறு

Question 4.
”மாபாரதம் தமிழ்ப்படுத்தும் மதுராபுரிச் சங்கம் வைத்தும் என்று குறிப்பிடும் செப்பேட்டுக் குறிப்பு
(அ) உத்திரமேரூர்
(ஆ) மண்டகப்பட்டு
(இ) சின்னமனூர்
(ஈ) ஆதிச்சநல்லூர்
Answer:
(இ) சின்னமனூர்

Question 5.
“தர்க்கத்திற்கு அப்பால் ” சிறுகதை அமைந்த தொகுப்பு…….
(அ) ரிஷிமூலம்
(ஆ) யுகசந்தி
(இ) குருபீடம்
(ஈ) ஒரு பிடி சோறு
Answer:
(ஆ) யுகசந்தி

Question 6.
மணிப்பால் பொறியியல் கல்லூரியில் பேராசிரியர்…….
(அ) ஜெயச்சந்திரன்
(ஆ) வேணுகோபாலன்
(இ முத்தையா
(ஈ) சாத்தப்பன்
Answer:
(ஆ) வேணுகோபாலன்

Question 7.
வேற்றுமைத் தொகையில் வேற்றுமை உருபு……………….. வரும்.
(அ) வெளிப்படையாக
(ஆ) மறைவாக
(இ முதலில்
(ஈ) கடைசியில்
Answer:
(ஆ) மறைவாக

Question 8.
செய்குதம்பிப் பாவலரின் அனைத்து நூல்களும்………… ஆக்கப்பட்டுள்ளது.
(அ) அரசுடைமை
(ஆ) மக்களுடைமை
(இ) நாட்டுடைமை
(ஈ) பொதுவுடைமை
Answer:
(ஈ) பொதுவுடைமை

Question 9.
‘மலர்க்கை என்பதனை உருவகமாக்கினால் …….. ………. என வரும்.
(அ) பண்புத்தொகை
(ஆ) உம்மைத்தொகை
(இ) வினைத்தொகை
(ஈ) அன்மொழித்தொகை
Answer:
(இ) வினைத்தொகை

Question 10.
மூன்று காலங்களுக்கும் பொருந்துமாறு அமைவது…… …..ஆகும்.
(அ) பண்புத்தொகை
(ஆ) உம்மைத்தொகை
(இ) வினைத்தொகை
(ஈ) அன்மொழித்தொகை
Answer:
(இ) வினைத்தொகை

Question 11.
முல்லைப்பாட்டு ………. பாவால் இயற்றப்பட்டது.
(அ) வெண்பா
(ஆ) ஆசிரியப்பா
(இ) வஞ்சிப்பா
(ஈ) கலிப்பா
Answer:
(ஆ) ஆசிரியப்பா

பாடலைப் படித்துப் பின்வரும் வினாக்களுக்கு (12, 13, 14, 15) விடை தருக.
நின்று காவல் நெறி பூண்டு நெறியல்லது நினையாது தந்தையில்லோர் தந்தையாகியுந் தாயரில்லோர் தாயராகியும் மைந்தரில்லொரு மைந்தராகியும் மன்னுயிர் கட்குயிராகியும் விழிபெற்ற பயனென்னவும் மெய்பெற்ற அருளென்னவும்

12. இப்பாடலில் வந்துள்ள எதுகையை குறிப்பிடுக.
(அ) தந்தை , மைந்தர்
(ஆ) நின்று, நினையாது
(இ) மைந்தரில், மைந்தரால்
(ஈ) தயாரிகி, கட்குயிராகி
Answer:
(அ) தந்தை , மைந்தர்

Question 13.
இப்பாடலில் இடம் பெற்ற உறவு முறை யாது?
(அ) மாமன், அத்தை
(ஆ) சித்தன், சித்தி
(இ தந்தை, தாய்
(ஈ) பாட்டி, தாத்தா
Answer:
(இ தந்தை, தாய்

Question 14.
மைந்தரில்லாத – பிரித்து எழுதுக.
(அ) மை + தரி + இல்லாத
(ஆ) மைந்து + இல்லாத
(இ) மைந்தர் + இல்லாத
(ஈ) மைந்தரி + இல்லாத
Answer:
(இ) மைந்தர் + இல்லாத

Question 15.
இப்பாடல் இடம் பெற்ற நூல் எது?
(அ) மெய்க்கீர்த்தி
(ஆ) சிலப்பதிகாரம்
(இ) காலக்கணிதம்
(ஈ) நீதி வெண்பா
Answer:
(அ) மெய்க்கீர்த்தி

பகுதி – II (மதிப்பெண்க ள்: 18)

பிரிவு – 1

எவையேனும் நான்கு வினாக்களுக்கு மட்டும் குறுகிய விடையளிக்க.
21 ஆவது வினாவிற்குக் கட்டாயமாக விடையளிக்க வேண்டும். [4×2 = 8]

Question 16.
விடைக்கேற்ற வினா அமைக்க.
(அ) கண்ணதாசனின் கவிதைத் தொகுப்பில் இடம் பெற்ற பகுதியின் பெயர் காலக்கணிதம் ஆகும்.
(ஆ) நூலின் பயன். அறம், பொருள், இன்பம், வீடு என்ற நான்கு பயனுக்காக இருத்தல் வேண்டும்.
Answer:
விடை:
(அ) கண்ணதாசனின் கவிதைத் தொகுப்பில் இடம் பெற்ற பகுதியின் பெயர் என்ன?
(ஆ) நூலின் பயன் எத்தகையது?

Question 17.
கவிஞர் சச்சிதானந்தனின் தமிழ்ப்பசியை எழுதுக.
Answer:
‘சாகும் போதும் தமிழ் படித்துச் சாக வேண்டும்” என்றார். “சாம்பலும் தமிழ்மணந்து வேகவேண்டும்” என்பதே கவிஞர் சச்சிதானந்தனின் தமிழ்ப்பசி ஆகும்.

Question 18.
பாரதியார் எவ்வாறெல்லாம் பாராட்டப்பட்டார்?
Answer:
பாரதியார் ‘நீடுதுயில் நீக்கப் பாடிவந்த நிலா’, ‘சிந்துக்குத் தந்தை’ ‘பாட்டுக்கொரு புலவன்’
எனப் பாராட்டப்பட்டார்.

Question 19.
“கழிந்த பெரும் கேள்வியினான் எனக் கேட்டு முழுது உணர்ந்த கபிலன் தன் பால்
Answer:
பொழிந்த பெரும் காதல் மிகு கேண்மையினான் இடைக்காட்டுப் புலவன் தென் சொல்” – இவ்வடிகளில் கழிந்த பெரும் கேள்வியினான் யார்? காதல்மிகு கேண்மையினான் யார்? இவ்வடியில் கழிந்த பெரும் கேள்வியினான். குசேல பாண்டியன் காதல்மிகு கேண்மையினான். இடைக்காடனார்.

Question 20.
தோற்பாவைக் கூத்தின் வேறு பெயர்கள் யாவை?
Answer:
தோற்பாவைக் கூத்தின் வேறு பெயர்கள்

• கையுறைப் பாவைக் கூத்து
• பொம்மலாட்டம் ஆகும்.

Question 21.
‘இன்மையின்’ எனத் தொடங்கும் குறள் எழுதுக.
Answer:
இன்மையின் இன்னாத தியாதெனின் இன்மையின் இன்மையே இன்னா தது

பிரிவு – 2

எவையேனும் ஐந்து வினாக்களுக்கு மட்டும் குறுகிய விடையளிக்க. [5×2 = 10]

Question 22.
உவமையைப் பயன்படுத்தி சொற்றொடர் அமைக்க
கண்ணினைக் காக்கும் இமை போல?
Answer:
என் தாயைக் கண்ணினைக் காக்கும் இமை போல பாதுகாத்தேன்.

Question 23.
பாடலில் இடம் பெற்றுள்ள தமிழ்ப் புலவர்களின் பெயர்களைக் கண்டறிந்து எழுதுக.
Answer:
”கம்பனும் கண்டேத்தும் உமறுப் புலவரை எந்தக்
கொம்பனும் பணியும் அறம்பாடுஞ் ஜவாது ஆசுகவியை
காசிம் புலவரை, குணங் குடியாரை சேகனாப் புலவரை
செய்குதம்பிப் பாவலரைச் சீர்தமிழ் மறக்காதன்றோ ”

• கம்பன்
• உமறுப்புலவர்
• ஆசுகவி
• காசிம் புலவர்
• குணங்குடியார்
• செய்குதம்பிப்புலவர்

Question 24.
அறுசுவை – தொகைச் சொல்லைப் விரித்து எழுதி தமிழ் எண்ணுரு தருக.
Answer:
இனிப்பு, புளிப்பு, கசப்பு, உவர்ப்பு, துவர்ப்பு, கார்ப்பு – (சா)

Question 25.
கலைச்சொற்கள் தருக.
Answer:

• Devotional Literature – பக்தி இலக்கியம்
• Folk Literature – நாட்டுப்புற இலக்கியம்

Question 26.
கீழ்க்காணும் மரபுத் தொடருக்கான பொருளறிந்து தொடர் அமைத்து எழுதுக.
Answer:
ஆறப்போடுதல் எந்த ஒரு பிரச்சனை வந்தாலும் முடிவுகளை உடனே எடுக்கக்கூடாது. ஆறப்போடுதல் வேண்டும்.

Question 27.
பொருத்தமான நிறுத்தக்குறிகளை இடுக.
Answer:
கடுகைத் துளைத்து ஏழ்கடலைப் புகட்டிக் குறுகத்தறித்த குறள் என இடைக்காடனார் கூறினார் போலும்.

விடை : ”கடுகைத் துளைத்து ஏழ்கடலைப் புகட்டிக் குறுகத்தறித்த குறள்’ என இடைக்காடனார் கூறினார் போலும்.

Question 28.
உரைத்த – பகுபத உறுப்பிலக்கணம் தருக.
Answer:
உரைத்த = உரை + த் + த் + அ
உரை – பகுதி
த் – சந்தி
த் – இறந்த கால இடைநிலை
அ – பெயரெச்ச விகுதி

பகுதி – III (மதிப்பெண்கள்: 18)

பிரிவு – 1

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க. [2×3 = 6]

Question 29.
படங்கள் வெளிப்படுத்தும் நிகழ்த்துக்கலை குறித்து இரண்டு வினாக்களையும் அவற்றுக்கான விடைகளையும் எழுதுக.

Answer:
பரதநாட்டியம் :
1. பரதநாட்டிய வளர்ச்சிப் பற்றி எழுதுக.
பழமையான கலைகளுள் ஒன்றாகும் பெண்கள் மட்டும் ஆடும் கலையாக இருந்தது. இன்று ஆண்களும், பெண்களும் ஆடும் கலையாக வளர்ச்சி அடைந்துள்ளது.

2. பரத நாட்டியம் எவ்வகை கலையாகக் கருதப்படுகிறது?
இறைவனுக்கு உகந்த கலையாகப் பரதநாட்டியம் கருதப்படுகிறது.

கரகம் :
1. கரகம் என்பது எத்தகைய நடனம்?
கரகாட்டம் என்பது கிராமிய நடனம் ஆகும். ஆணும், பெண்ணும் சேர்ந்து ஆடும் ஆட்டம் ஆகும்.

2. கரகாட்டத்தின் வேறு பெயர் என்ன?
கரகாட்டம் கரகம், கும்பாட்டம் என்றும் அழைக்கப்படுகிறது.

Question 30.
பல்துறை வளர்ச்சியின் மொழிபெயர்ப்பின் பங்கு என்ன?
Answer:

• மொழிபெயர்ப்பு இல்லை எனில் உலகை எல்லாம் வலையாகப் பிடித்திருக்கிற ஊடகத்தின் வளர்ச்சி இல்லை.
• தொலைக்காட்சி, வானொலி, திரைப்படம், இதழ்கள் போன்ற ஊடகங்கள் மொழிபெயர்ப்பால் தான் வளர்ச்சி பெறுகின்றன.
• விளம்பர மொழிக்கு மொழிபெயர்ப்பு தேவைப்படுகிறது.
• திரைப்படங்கள் தொலைக்காட்சித் தொடர்கள் ஆகியன வேற்று மொழிமாற்றம் செய்யப்பட்டு அனைத்து மொழி பேசும் மக்களையும் அடைகின்றன.
• இதனால் புதுவகையான சிந்தனைகள் மொழிக்கூறுகள் பரவுகின்றன.

Question 31.
உரைப்பத்தியைப் படித்து வினாக்களுக்கு விடை தருக.
Answer:
செயற்கை நுண்ணறிவு என்பது ஒரு மென்பொருள் அல்லது கணினிச் செயல்திட்ட வரைவு எனலாம். அது தானாகக் கற்றுக் கொள்ளக்கூடியது. இந்த அறிவைக் கொண்டு தனக்கு வரும் புதிய புதிய சூழ்நிலைகளில் மனிதரைப்போல, தானே முடிவெடுக்கும் திறனுடையது.

ஒளிப்படங்கள், எழுத்துக்கள், காணொலிகள், ஒலிகள் போன்றவற்றிலிருந்து கற்றுக்கொள்ளும் இயல்புடைய கனகா கானை மென்பொருளை ஆராய்ச்சியாளர் வடிவமைக்கிறார். அவ்வாறு கற்றுக் கொண்டதை அந்த இயந்திரம் தேவைப்படும் இடங்களில், தேவைப்படும் நேரங்களில் செயல்படுத்தும்.

செயற்கை நுண்ணறிவு பொதிந்த இயந்திரங்களுக்கு ஓய்வு தேவையில்லை; செயற்கை நுண்ணறிவால் பார்க்கவும், கேட்கவும், புரிந்துக்கொள்ளவும் முடியும் என்பதே அதன் சிறப்பு. மனிதனால் முடியும் செயல்களையும் அவன் கடினம் என்று கருதும் செயல்களையும் செய்யக்கூடியது செயற்கை நுண்ணறிவு.

(அ) செயற்கை நுண்ணறிவு என்பது யாது?
Answer:
செயற்கை நுண்ணறிவு என்பது ஒரு மென்பொருள் அல்லது கணினிச் செயல்திட்ட வரைவு எனலாம்.

(ஆ) மென்பொருளை ஆராய்ச்சியாளர்கள் எவ்வாறு வடிவமைக்கின்றனர்?
ஒளிப்படங்கள், எழுத்துக்கள், காணொலிகள், ஒலிகள் போன்றவற்றிலிருந்து கற்றுக்கொள்ளும் வண்ணம் வடிவமைக்கின்றனர்.

(இ) செயற்கை நுண்ணறிவின் சிறப்பு யாது?
Answer:
பார்க்கவும், கேட்கவும், புரிந்துக் கொள்ளவும் முடியும்.

பிரிவு – 2

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க.
34 ஆவது வினாவிற்குக் கட்டாயமாக விடையளிக்க வேண்டும். [2 x 3 = 6]

Question 32.
மாளாத காதல் நோயாளன் போல் என்னும் தொடரிலுள்ள உவமை சுட்டும் செய்தியை விளக்குக.
Answer:
மருத்துவர் உடலில் ஏற்பட்ட புண்ணைக் கத்தியால் அறுத்துச் சுட்டாலும் அது நன்மைக்கே என்று உணர்ந்து நோயாளி அவரை நேசிப்பார் வித்துவக் கோட்டில் எழுந்தருளியிருக்கும் அன்னையே! அதுபோன்று நீ உனது விளையாட்டால் நீங்காத துன்பத்தை எனக்குத் தந்தாலும் உன் அடியவனாகிய நான் உன் அருளையே எப்பொழுதும் எதிர்பார்த்து வாழ்கின்றேன்.

உன் காற்றே உடலுக்கு நல்லது. தூய்மையானது. அதை மனிதர்கள் புரிந்து கொள்ள வேண்டும்.

Question 33.
முகம்மதுரஃபி ஆசிரியர் குறிப்பு வரைக?
Answer:

• முகம்மது ரஃபி என்னும் இயற்பெயரைக் கொண்ட நாகூர் ரூமி தஞ்சை மாவட்டத்தில் பிறந்தவர்.
• இவர் எண்பதுகளில் கணையாழி இதழில் எழுதத் தொடங்கியவர்.
• கவிதை, குறுநாவல், சிறுகதை, மொழிபெயர்ப்பு எனப் பலதளங்களில் இவர் தொடர்ந்து இயங்கி வருபவர்.
• மீட்சி, சுபமங்களா, புதிய பார்வை, குங்குமம், கொல்லிப்பாவை, இலக்கிய வெளிவட்டம், குமுதம் ஆகிய இதழ்களில் இவரது படைப்புகள் வெளியாகியுள்ளன.
• இதுவரை நதியின் கால்கள், ஏழாவது சுவை, சொல்லாத சொல் ஆகிய மூன்று கவிதைத் தொகுதிகள் வெளியாகியுள்ளன.
• மொழிபெயர்ப்புக் கவிதைகள், சிறுகதைத்தொகுதிகள் ஆகியவற்றுடன் கப்பலுக்குப் போன மச்சான்’ என்னும் நாவலையும் படைத்துள்ளார்.

Question 34.
அடிபிறழாமல் எழுதுக.
(அ) “அருளைப் பெருக்கி ” எனத் தொடங்கும் ‘நீதிவெண்பா ‘ பாடல்.
Answer:
அருளைப் பெருக்கி அறிவைத் திருத்தி
மருளை அகற்றி மதிக்கும் தெருளை
அருத்துவதும் ஆவிக்கு அருத்துணையாய்
இன்பம் பொருத்துவதும் கல்வியென்றே போற்று
– கா.ப. செய்குதம்பிப் பாவலர்

(அல்லது)

(ஆ ) “அன்னை மொழியே” எனத் தொடங்கும் பாடல்.
Answer:
அன்னை மொழியே ! அழகார்ந்த செந்தமிழே!
முன்னைக்கும் முன்னை முகிழ்த்த நறுங்கனியே!
கன்னிக் குமரிக் கடல் கொண்ட நாட்டிடையில்
மன்னி அரசிருந்த மண்ணுலகப் பேரரசே!
தென்னன் மகளே ! திருக்குறளின் மாண்புகழே!
இன்னறும் பாப்பத்தே! எண்தொகையே நற்கணக்கே!
மன்னுஞ் சிலம்பே! மணிமே கலைவடிவே!
முன்னும் நினைவால் முடிதாழ வாழ்த்துவமே!
– பாவலரேறு பெருஞ்சித்திரனார்

பிரிவு – 3

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க. [283 = 6]

Question 35.
சிறுபொழுது ஆறு கூறுகள் யாவை?
Answer:

• காலை – காலை 6 மணி முதல் 10 மணி வரை
• நண்பகல் – காலை 10 மணி முதல் 2 மணி வரை
• எற்பாடு – பிற்பகல் 2 மணி முதல் 6 மணி வரை
• லை மாலை 6 மணி முதல் இரவு 10 மணி வரை
• யாமம் – இரவு 10 மணி முதல் 2 மணி வரை
• வைகறை – இரவு 2 மணி முதல் காலை 6 மணி வரை

Question 36.
‘பல்லார் பகைகொளலின் பத்தடுத்த தீமைத்தே நல்லார் தொடர்கை விடல்’ இக்குறட்பாவினை அலகிட்டு வாய்பாடு தருக.
Answer:

Question 37.
நிரல்நிறை அணி என்றால் என்ன? சான்றுடன் விளக்குக.
Answer:
நிரல் = வரிசை ; நிறை = நிறுத்துதல் பார்த்த சொல்லையும் பொருளையும் வரிசையாக நிறுத்தி அவ்வரிசைப்படியே இணைத்துப் பொருள் கொள்வது நிரல்நிறை அணி எனப்படும்.

(எ.கா.) அன்பும் அறனும் உடைத்தாயின் இல்வாழ்க்கை
பண்பும் பயனும் அது.

பாடலின் பொருள்: இல்வாழ்க்கை அன்பும், அறமும் உடையதாக விளங்குமானால், அந்த வாழ்க்கையின் பண்பும் பயனும் அதுவே ஆகும். அணிப்பொருத்தம். இக்குறளில் அன்பும் அறனும் என்ற சொற்களை வரிசையாக நிறுத்தி, பண்பும் பயனும் என்ற சொற்களை முறைப்படக் கூறியுள்ளமையால், இது நிரல் நிறை அணி ஆகும்.

பகுதி – IV (மதிப்பெண்கள்: 25)

அனைத்து வினாக்களுக்கும் விடையளிக்க. [5 x 5 = 25]

Question 38.
(அ) முல்லைப் பாட்டில் உள்ள கார்காலச் செய்திகளை விவரித்து எழுதுக.
Answer:
முன்னுரை:
தமிழர்கள் இயற்கையோடு இயைந்த வாழ்வைக் கொண்டிருந்தனர் மழைக்காலத்தில் அவர்கள் வாழ்வை எதிர்கொள்கிற இயல்பு இலக்கியத்தில் பதிவு செய்யப்பட்டிருக்கிறது.

மழை பொழியும் காட்சி :
வலம்புரிச்சங்கு பொறித்த கைகளையுடைய திருமால், குறுகிய வடிவம் கொண்டு மாவலி மன்னன் நீர் வார்த்துத் தரும்பொழுது, மண்ணுக்கும் விண்ணுக்குமாகப் பேருருவம் எடுத்து உயர்ந்து நிற்பது போன்றுள்ளது மழைமேகம். அம்மேகம், ஒலிக்கும் கடலின் குளிர்நீரைப் பருகிப் பெருந்தோ கொண்டு , வலமாய் எழுந்து, மலையைச் சூழ்ந்து, விரைந்த வேகத்துடன் பெருமழையைப் பொழிகிறது.

முதிய பெண்கள் நற்சொல் கேட்டு நின்ற காட்சி:
முதிய பெண்கள் மிகுந்த காவலையுடைய ஊர்ப்பக்கம் சென்றனர். யாழிசை போன்று ஒலிக்கும் வண்டுகள் சூழ்ந்து ஆரவாரிக்கும் நறுமணம் கொண்ட அரும்புகள், அந்த மலர்ந்த முல்லைப் பூக்களோடு நாழியில் கொண்டு வந்த நெல்லையும் சேர்த்துத் தெய்வத்தின் முன் தூவினர். பிறகு தெய்வத்தைத் தொழுது தலைவிக்காக நற்சொல் கேட்டு நின்றனர்.

முதுபெண்டிர் விரிச்சி கேட்டு நின்ற காட்சி:
சிறு தாம்புக் கயிற்றால் கட்டப்பட்ட இளங்கன்று பசியால் வாடிக்கொண்டிருந்தது. அதன் வருத்தத்தை ஓர் இடைமகள் கண்டாள். குளிர் தாங்காமல் கைகளைக் கட்டியபடி நின்ற அவள் புல்லை மேய்ந்து உன் தாய்மார் வளைந்த கத்தியை உடைய கம்பைக் கொண்ட எம் இடையர் ஓட்டிவர இப்போது வந்துவிடுவர், வருந்தாதே என்றாள்.

இது நல்ல சொல் எனக்கொண்டு முதுபெண்கள் தலைவியிடம் நற்சொல்லை நாங்கள் கேட்டோம் என்று கூறினர். இவ்வாறு தலைவன் வருகை குறித்து முதுபெண்டிர் விரிச்சி கேட்டு நின்றனர்.

முடிவுரை:
முல்லை நிலத்தின் மழைப்பொழிவையும், முல்லை மலரும் நெல்லும் தூவி முதுபெண்டிர் தெய்வத்தை வழிபட்டதையும், விரிச்சி கேட்டதையும் முல்லைப்பாட்டில் நப்பூதனார் படம் பிடித்துக் காட்டியுள்ளார்.

(அல்லது)

(ஆ) காலக்கணிதம் கவிதையில் பொதிந்துள்ள நயங்களைப் பாராட்டி எழுதுக
Answer:
கவிஞன் யானோர் காலக் கணிதம்
கருப்படு பொருளை உருப்பட வைப்பேன்!
புவியில் நானோர் புகழுடைத் தெய்வம்
பொன்னினும் விலைமிகு பொருளென் செல்வம்!
இவைசரி யென்றால் இயம்புவதென் தொழில்
இவைதவ றாயின் எதிர்ப்பதென் வேலை!
ஆக்கல் அளித்தல் அழித்தல் இம் மூன்றும்
அவனும் யானுமே அறிந்தவை, அறிக!

– கண்ணதாசன்

கருத்து:
நான் தான் காலக் கணிதன் கருப்படும் பொருளை உருப்பட வைப்பேன்! புவியில் நல்லவர்கள் பலபேர் இருக்கின்றனர். பொன்னும் விலைமிகு பொருளும் இருக்கிறது. அது செல்வம், இதுசரி, இது தவறு என்று சொல்வது என் வேலை செய்வது தவறாயின் எதிர்ப்பது என் வேலை சரி என்றால் புகழ்வது என் தொழில். ஆக்கல் காத்தல், அழித்தல் இம்மூன்றும் இறைவனும் நானும் மட்டுமே அறிந்த தொழில்களாகும்.

எதுகை: செய்யுளின் இரண்டாம் எழுத்து ஒன்றிவரத் தொடுப்பது எதுகை
கவிஞன், புவியில்

மோனை: செய்யுளில் முதல் எழுத்து ஒன்றிவரத் தொடுப்பது மோனை.
கவிஞன், காலம், கணிதம், கருப்படு

முரண் : சரி தவறு x ஆக்கல் x அழித்தல்

சொல் நயம்: கவிஞன் யானோர் காலக்
கணிதம் கருப்படு பொருளை உருப்பட வைப்பேன்

என்ற சொற்றொடர்களை அமைத்துப் பாடலுக்குச் சிறப்புச் சேர்த்துள்ளார்.

(எ.கா.) தெய்வம் எனத் தன்னைக் கூறும் கவிஞர் புகழுடைத் தெய்வம் என்ற சொற்றொடரைக் கையாளும் நயம் படித்து இன்புறத்தக்கது. பொருள் நயம்: ஆக்கல் அளித்தல் அழித்தல் இம்மூன்றும் அவனும் யானுமே அறிந்தவை என்றும் ஆழ்ந்த பொருள் சுவை உடையது (எ.கா) தன் செல்வம் எது எனக் கூற வந்த கவிஞர் பொன் விலை உயர்ந்தது. அதைக் காட்டிலும் விலை உயர்ந்த கவிதைப்பொருளே என் செல்வம் எனக் கூறியிருக்கும்.

இக்கவிதையின் பொருள் நயம் போற்றுதற்குரியது.

Question 39.
(அ) உங்கள் தெருவில் மின்விளக்குகள் பழுதடைந்துள்ளன. அதனால் இரவில் சாலையில் நடந்து சொல்வோருக்கும் வாகன ஓட்டிகளுக்கும் ஏற்படும் இடையூறுகளைக் குறிப்பிட்டு, புதிய மின்விளக்குகள் பொருத்தும்படி மின்வாரிய அலுவலருக்குக் கடிதம் எழுதுக.
Answer:

விழுப்புரம்,
18.05.2019.

அனுப்புநர்
பொது மக்கள்,
பூந்தோட்டம்,
விழுப்புரம் – 05.

பெறுநர்
மின்வாரிய இயக்குநர்,
மின்வாரிய அலுவலகம்,
பூந்தோட்டம்,
விழுப்புரம்-05.
ஐயா,

பொருள்: தெரு விளக்கு பழுதுநீக்கித் தருமாறு விண்ணப்பம் அளித்தல் – சார்பு. வணக்கம். எங்கள் பகுதியில் ஏறக்குறைய மூவாயிரம் பேர் வாழ்கிறார்கள். தெருக்களில் விளக்குகள் ஒளி வழங்குவது இல்லை. அதனால் தெருக்களில் நாய்கள் படுத்து உறங்குவது தெரியாமல் மிதித்து விடுகின்றனர். அதனால் நாய்கள் தெரு வழியே செல்வோரைக் கடித்துவிடுகின்றன. நாய் கடியினால் வருந்துவோர்களின் எண்ணிக்கை மிகுதியாக உள்ளது.

தெரு விளக்குகள் இயங்காமையால் தெருவில் நடந்து செல்வோர், விபத்துக்கும் ஆளாகின்றனர். தவிர முகமூடிக் கொள்ளையர் தொடர்ச்சியாக வீடுகளில் புகுந்து திருடிச் செல்கின்றனர். உயிர்க் கொலையும் செய்கின்றனர். தெரு விளக்குகளை விரைவாகச் சீர்செய்து எங்கள் துன்பத்தைப் போக்க ஆவன செய்யுமாறு அன்புடன் வேண்டுகிறோம்.

நன்றி

இங்ஙனம்,
உங்கள் உண்மையுள்ள,
பொதுமக்கள்.

பூந்தோட்டம்,
4.4.2019.
உறைமேல் முகவரி

பெறுநர்
மின்வாரிய இயக்குநர்,
மின்வாரிய அலுவலகம்,
பூந்தோட்டம்,
விழுப்புரம்-05.

அல்லது)

(ஆ) நாளிதழ் ஒன்றின் பொங்கல் மலரில், உழவுத் தொழிலுக்கு வந்தனை செய்வோம்’ என்ற உங்கள் கட்டுரையை வெளியிட வேண்டி, அந்நாளிதழ் ஆசிரியருக்குக் கடிதம் எழுதுக.
Answer:
அனுப்புநர் –
தெ. தண்டபாணி,
35, மேற்கு மாடவீதி,
மதுரை – 625001.

பெறுநர்
தினமணி ஆசிரியர்,
தினமணி அலுவலகம்,
எக்ஸ்பிரஸ் எஸ்டேட்,
மதுரை – 625003.

பொருள்: எனது கட்டுரையை வெளியிட வேண்டி விண்ணப்பம். ஐயா,

வணக்கம்.

இந்த ஆண்டு பொங்கல் விழா எங்கள் ஊரில் மிகச்சிறப்பாக நடைப்பெற்றது. நான்கு நாட்கள் நடைபெற்ற நிகழ்வுகள் அனைவரையும் மிகவும் கவர்ந்தன. அதன் தொடர்பாக நான் “உழவுக்கும் தொழிலுக்கும் வந்தனை செய்வோம்” என்னும் தலைப்பில் ஒரு கட்டுரை எழுதி இருக்கிறேன். அந்தக் கட்டுரையைத் தங்களின் நாளிதழில் வெளியிடுமாறு கேட்டுக் கொள்கிறேன்.

நன்றி,
இடம் : மதுரை
தேதி 2.04.2019

இங்ஙனம்,
தங்கள் உண்மையுள்ள,
தெ. தண்டபாணி

குறிப்பு
இத்துடன் கட்டுரை இணைக்கப்பட்டுள்ளது.
உறைமேல் முகவரி

பெறுநர்
தினமணி ஆசிரியர்,
தினமணி அலுவலகம்,
எக்ஸ்பிரஸ் எஸ்டேட்,
மதுரை – 625003.

Question 40.
படம் உணர்த்தும் கருத்தை நயமுற நான்கு தொடர்களில் எழுதுக.

Answer:
வறியவர்க்கு ஒன்று ஈவது ஈகை
மறித்துக் கொடுக்காமல் தடுப்பது தீமை
முடிந்ததைக் கொடுப்பது மேதை
முடிந்ததைத் தடுப்பது பேதை
வாடி நிற்கும் வறியவர்க்குக் கொடுப்பது புகழ்
கொடுப்பதைத் தடுத்து நிறுத்துவது இகழ்
பாத்தாள் –

Question 41.
கீழ்க்காணும் படிவத்தை நிரப்புக.
Answer:

நூலக உறுப்பினர் படிவம்

மதுரை மாவட்ட நூலக ஆணைக்குழு
மைய / கிளை / ஊர்ப்புற நூலகம் மைய நூலகம்.

உறுப்பினர் சேர்க்கை அட்டை)
அட்டை எண் – உறுப்பினர் எண் 567

• பெயர் – கந்தன்
• தந்தை பெயர் – ஆறுமுகம்
• பிறந்த தேதி – 06.06.2005
• வயது – 14
• படிப்பு – பத்தாம் வகுப்பு
• தொலைபேசி எண் – 98678 64590
• முகவரி – 35 அம்மன் கோயில் தெரு

(அஞ்சல் குறியீட்டு எண்ணுடன்) 5 வது தெரு, மேலவீதி, மதுரை – 625002.

அ. கந்தன் நூலகத்தில் உறுப்பினராகப் பதிவு செய்ய இத்துடன் காப்புத்தொகை ரூ 100 சந்தா தொகை ரூ. 100 ஆக மொத்தம் ரூ 200 ரொக்கமாகச் செலுத்துகிறேன். நூலக நடைமுறை மற்றும் விதிகளுக்குக் கட்டுப்படுகிறேன் என உறுதியளிக்கிறேன்.

இடம் : மதுரை
நாள் : 24.5.2019

தங்கள் உண்மையுள்ள
அ. கந்தன்

திரு திருமதி செல்வி / செல்வன் அ. கந்தன் அவர்களை எனக்கு நன்கு தெரியும் எனச் சான்று அளிக்கிறேன்.

அலுவலக முத்திரை

தீபா
பிணைப்பாளர் கையொப்பம்
(பதவி மற்றும் அலுவலகம்)
(மாநில மைய அரசு அதிகாரிகள், கல்லூரி முதல்வர்கள் பேராசிரியர்கள், உயர் / மேல்நிலைப்பள்ளி தலைமை ஆசிரியர்கள், சட்டமன்ற / நாடாளுமன்ற
உறுப்பினர்கள், நகராட்சி/மாநகராட்சி ஒன்றிய பேரூராட்சி உறுப்பினர்கள்)

Question 42.
அ நாம் எப்போதும் ஒரே மனநிலையில் இருப்பதில்லை. நம்மைச் சுற்றி நிகழும் செயல்களால் நாம் அலைக்கழிக்கப்படுகிறோம். உடன்பயில்பவருடனோ, உடன் பிறந்தவருடனோ எதிர்பாராமல் சச்சரவு ஏற்படுகிறது……. இந்தச் சமயத்தில் சினம் கொள்ளத் தக்க சொற்களைப் பேசுகிறோம்; கேட்கிறோம், கைகலப்பில் ஈடுபடுகிறோம். இதுகாறும் கற்ற அறங்கள் நமக்குக் கைகொடுக்க வேண்டாமா? மாணவ நிலையில் நாம் பின்பற்ற வேண்டிய அறங்களும் அதனால் ஏற்படும் நன்மைகளையும் வரிசைப்படுத்தி எழுதுக.
Answer:

(அல்ல து)
மொழிபெயர்க்க.
Among the five geographical divisions of the Tamil country in Sangam literature, the Marutam region was the fit for cultivation, as it had the most fertile lands. The property of a farmer depended on getting the necessary sunlight, seasonal rains and the fertility of the soil. Among these elements of nature, sunlight was considered indispensable by the ancient Tamils.

விடை: சங்க இலக்கியங்களில் இடம் பெற்ற தமிழ்நாட்டின் 5 புவியியல் பாகுபாட்டின்படி, மருத நிலப் பகுதியே பயிரிடுவதற்குச் செழுமையான பகுதியாகக் கருதப்பட்டது. விவசாயியின் சொத்து அங்கு கிடைக்கும் வெயில், பருவ மழை மற்றும் நிலத்தின் செழுமையைச் சார்ந்திருந்தது. இயற்கையில் கிடைக்கும் மூலக்கூறுகளில் சூரிய ஒளியே இன்றியமையாததாகப் பழந்தமிழர்களால் கருதப்பட்டது.

பகுதி – V (மதிப்பெண்கள் : 24) 

அனைத்து வினாக்களுக்கும் விரிவாக விடையளிக்க. [3 x 8 = 24]

Question 43.
(அ) மலர்ந்தும் மலராத பாதிமலர் போல
Answer:
வளரும் விழி வண்ணமே – வந்து
விடிந்தும் விடியாத காலைப் பொழுதாக
விளைந்த கலை அன்னமே
நதியில் விளையாடி கொடியில் தலைசீவி
நடந்த இளந் தென்றலே – வளர்
பொதிகை மலை தோன்றி மதுரை நகர் கண்டு
பொலிந்த தமிழ் மன்றமே –
கவிஞர் கண்ணதாசனின் இப்பாடலில் தவழும் காற்றையும் கவிதை நயத்தையும் பாராட்டி உரைசெய்க.

முன்னுரை:
இயற்கையின் கூறுகளில் காற்றின் பங்கு கூடுதலானது எங்கும் நிறைந்திருப்பது உயிர்களின் உயிர் மூச்சுக் காற்றைக் கண்களால் காண முடியாது. மெய்யால் மட்டுமே உணரக்கூடியது காற்று நம்மை மெல்லத் தொட்டுச் சென்றால் தென்றல் எனப்படுகிறது. கோவா மக்களை கரைக்கதை மாதிரி வினாத்தாள் -4 – 71

தென்றல் காற்று :
தெற்கிலிருந்து வீசுவதால் தென்றல் காற்று எனப்படுகிறது. மரம், செடி, கொடி, ஆறு, மலை, பள்ளத்தாக்கு எனப் பல தடைகளைத் தாண்டி வருவதால் வேகம் குறைந்து இதமான இயல்பு கொள்கிறது. இந்த மென்காற்றை இளந்தென்றல் என்பர்.

இலக்கியத்தில் தென்றல்:
தென்றல் காற்று பலவித மலர்களின் நறுமணத்தை அள்ளி வரும் பொழுது கூடவே வண்டுகளையும் அழைத்து வருவதால் இளங்கோவடிகள் ” வண்டொடு புக்க மணவாய்த் தென்றல்” என நயம்பட உரைக்கிறார். பலபட்டடைச் சொக்கநாதப் புலவர் எழுதிய பத்மகிரிநாதர் தென்றல் விடு தூது என்னும் சிற்றிலக்கியத்தில்.

“நந்தமிழும் தண்பொருநை நன்னதியும் சேர் பொருப்பிற்
செந்தமிழின் பின்னுதித்த தென்றலே”

என தென்றலை பெண்ணொருத்தி அன்போடு அழைக்கிறாள்.

கண்ணதாசனின் கவிதை நயம்:
முழுவதும் மலராத மலர் மணத்தையும் அழகையும் கூட்டி வைத்திருக்கும். அம்மலரைப்போல வளரும் கண்ணின் வண்ணமே எனவும் விடிந்தும் விடியாத குளிர்ந்த காலை நேரத்தில் தோன்றிய கலை அன்னமே எனவும், நதியில் விளையாடி கொடிகளில் பாய்ந்து தலை சீவி தவழ்ந்து நடந்து வருகின்ற இளம் தென்றலே எனவும், பொதிகை மலையில் அகத்தியரால் வளர்க்கப்பட்டு மதுரை தமிழ் சங்கங்களில் அழகாய் வளர்ந்த தமிழே எனவும் குழந்தையைக் கண்ணதாசன் பாடுகிறார்.

முடிவுரை:
இவ்வாறாக இலக்கியப் படைப்புகளிலும் திரையிசைப் பாடல்களிலும் தென்றல் காற்று இன்றளவும் நீங்கா இடம் பெற்றுள்ளது. மென்துகிலாய் உடல் வருடி மாயங்கள் செய்வது தென்றல் காற்றேயாகும்.

(அல்லது)

(ஆ) நிகழ்கலை வடிவங்கள் – அவை நிகழும் இடங்கள் – அவற்றின் ஒப்பனைகள் – சிறப்பும் பழைமையும் – இத்தகைய மக்கள் கலைகள் அருகிவருவதற்கான காரணங்கள் – அவற்றை வளர்த்தெடுக்க நாம் செய்ய வேண்டுவன – இவை குறித்து நாளிதழுக்கான தலையங்கம் எழுதுக. நிகழ்கலை வடிவங்கள் நிலைக்குமா?!
Answer:
கண்ணுக்குக் காட்சியையும் சிந்தைக்குக் கருத்தினையும் தருவன. கருத்துடன் கலைத்திறனை நோக்காகக் கொண்டு காலவெள்ளத்தைக் கடந்து நிற்பன ஆடல், பாடல், இசை , நடிப்பு ஒப்பனை உரையாடல் வழியாக மக்களை மகிழ்வடையச் செய்வன. சமூகப் பண்பாட்டுத்தளத்தின் கருத்துக் கருவூலமாக விளங்குவன. நுட்பமான உணர்வுகளின் உறைவிடமாக இருப்பன. அவை யாவை? அவை தாம் மக்கள் பண்பாட்டின் பதிவுகளான நிகழ்கலைகள்.

சிற்றூர் மக்களின் வாழ்வியல் நிகழ்வுகளில் பிரித்துப் பார்க்க இயலாக் கூறுகளாகத் திகழ்வை நிகழ்கலைகள். இவை மக்களுக்கு மகிழ்ச்சியெனும் கனி கொடுத்துக் கவலையைப் போக்குகின்றன. சமுதாய நிகழ்வுகளின் ஆவணங்களாகவும் செய்திகளைத் தரும் ஊடகங்களாகவும் திகழ்கின்றன. பழந்தமிழ் மக்களின் கலை, அழகியல், புதுமை ஆகியவற்றின் எச்சங்களை அறிவதற்குத் தற்காலத்தில் நிகழ்த்தப்படும் கலைகள் துணை செய்கின்றன.

நிகழ்கலைகள் ஊரக மக்களின் வாழ்வில் இரண்டறக் கலந்திருக்கின்றன. இவை கற்றோராலும் மற்றோராலும் விரும்பப்படும் கலைகளாக உள்ளன. உழைப்பாளிகளின் உணர்வுகளாக உள்ளன. மக்களின் எண்ண வெளிப்பாடாக, வாழ்க்கையைக் காட்டும் கண்ணாடியாக, மக்களின் சமய வழிபாட்டிலும் வாழ்வியல் நிகழ்வுகளிலும் பிரிக்க முடியாத பண்பாட்டுக் கூறுகளாக விளங்குகின்றன.

நிகழ்கலைகளை வளரச் செய்வோம். என்றும் அழியாமல் நிலைக்கச் செய்வோம்.

Question 44.
(அ) அன்னமய்யா என்னும் பெயருக்கும் அவரின் செயலுக்கும் உள்ள பொருத்தப்பாட்டினைக் கோபல்லபுரத்து மக்கள் கதைப்பகுதி கொண்டு விவரிக்க. கதைக்கரு : கிராமத்து மனிதர்கள் காட்டும் விருந்தோம்பல், பகிர்ந்து கொடுக்கிற நேயம்.
Answer:
கதைமாந்தர்கள் :

• சுப்பையா
• கிராமத்து மக்கள்
• அன்னமய்யா
• மணி

முன்னுரை: கிராமத்து வெள்ளந்தி மனிதர்கள் காட்டும் விருந்தோம்பல் இயல்பான வரவேற்பும் எளிமையான உணவும் பசித்த வேளையில் வந்தவர்களுக்குத் தம்மிடம் இருப்பதைப் பகிர்ந்து கொடுக்கிற மனித நேயம் ஆகியவற்றை இக்கதைப்பகுதி எடுத்துக் கூறுகிறது.

கிராமத்து காட்சி :
அதிகாலை நேரத்தில் பாச்சல் அருகு எடுத்து முடித்துவிட்டுக் காலைக் கஞ்சியைக் குடிக்க உட்காரும் வேளையில் அன்னமய்யா யாரோ ஒரு சன்னியாசியைக் கூட்டிக் கொண்டு வருவதைக் கண்டான் சுப்பையா வரட்டும் வரட்டும். ஒரு வயிற்றுக்குக் கஞ்சி ஊற்றி நாமும் குடிப்போம் என்றார். கொத்தாளி அந்தப் புஞ்சை சாலையோரத்தில் இருந்ததால் தேசாந்திரிகள் வந்து இவர்களிடம் தண்ணீரோ, கஞ்சியோ சாப்பிட்டு விட்டுப் போவது வழக்கம்.

அன்னமய்யா கண்ட காட்சி :
நடக்க முடியாமல் உட்கார்ந்து உட்கார்ந்து எழுந்திருந்து ஆயாசமாக மெதுவாக நடந்து வந்து தாடியும் அழுக்கு ஆடையும் தள்ளாட்டமுமாக நடந்து வந்து கொண்டிருந்தவனைப் பார்க்கும் போது வயோதிகனாகவும் சாமியாரைப்போலவும் எண்ண வைத்தது. தற்செயலாக இவனைக்கண்ட அன்னமய்யா அவன் அருகில் சென்று பார்த்த பிறகுதான் தெரிந்தது அவன் ஒரு வாலிபன் என்று, கால்களை நீட்டி புளிய மரத்தில் சாய்ந்து உட்கார்ந்திருந்த அவனை நெருங்கிப் பார்த்தபோது பசியால் அவன் முகம் வாடிப்போயிருந்தது.

அன்னமய்யாவின் செயல்:
பசியால் வாடிப்போயிருந்த அவன் முகத்தில் தீட்சணியம் தெரிந்தது தன்னைப் பார்த்து ஒரு நேசப்புன்னகை காட்டிய அந்த வாலிப மனிதனைப் பார்த்துக்கொண்டே நின்றான் அன்னமய்யா. குடிக்கக் கொஞ்சம் தண்ணீர் கிடைக்குமா? என்ற அவனைத் தன்னோடு மெதுவாக நடக்க வைத்து அழைத்துச் சென்றான் அன்னமய்யா.

அன்னமய்யாவின் விருந்தோம்பல் :
வேப்பமரத்தின் அடியில் ஏகப்பட்ட மண் கலயங்கள் இருந்தன. அதில் அன்னமய்யா ஒரு கலயத்தின் மேல் வைக்கப்பட்ட கல்லை அகற்றிச் சிரட்டையைத் துடைத்துச் சுத்தப்படுத்தி அந்த கலயத்தில் பதனமான வடித்த நீரை அவனிடம், உறிஞ்சி குடிங்க எனக் கொடுத்தான். உட்கார்ந்து குடிங்க என்று உபசரித்தான்.

பிறகு கலயத்தைச் சுற்றி ஆட்டியதும் தெளிவு மறைந்து சோற்றின் மகுளி மேலே வந்ததும் வார்த்துக் கொடுத்தான். பிறகு அன்னமய்யா அந்த புது ஆளைச் சுப்பையாவின் வயலுக்கு அழைத்துச் சென்று கம்மஞ்சோற்றைச் சாப்பிட வைத்தான். அந்த வாலிபன் அன்னமய்யா என்ற பெயரை மனசுக்குத் திருப்பித் திருப்பிச் சொல்லிப் பார்த்துக் கொண்டான். எவ்வளவு பொருத்தம் என்று நினைத்துக் கொண்டான்.

முடிவுரை:
வந்தவனுக்கு எப்படி ஒரு நிறைவு ஏற்பட்டதோ அதை விட மேலான ஒரு நிறைவு அன்னமய்யாவுக்கு ஏற்பட்டது. வயிறு நிறைந்ததும் தூங்கிவிடும் குழந்தையைப் பார்ப்பது போல அவனை ஒரு பிரியத்தோடு பார்த்துக் கொண்டிருந்தான் அன்னமய்யா

(அல்லது)

(ஆ) மங்கையராய்ப் பிறப்பதற்கே….. எனும் தலைப்பில் எம்.எஸ்.சுப்புலெட்சுமி பற்றிய உனது கருத்தை சுருக்கமாக எழுதுக.
Answer:
முகில் நாச்சி (எம். எஸ். சுப்புலட்சுமி):
எம். எஸ். சுப்புலட்சுமி தமிழரின் பெருமையை உலக அரங்கான ஐ.நா. அவையில் பரப்பும் வகையில் அங்குத் தமிழ்நாட்டின் செவ்வியல் இசையைப் பாடியவர், காற்றினிலே வரும் கீதமாய் மக்கள் மனதில் நீங்கா இடம் பெற்றவர், இசைப்பேரரசி என்று நேரு பெருமகனாரால் அழைக்கப்பட்டவர் எம். எஸ். சுப்புலட்சுமி.

தாழம்பூ குங்குமமிட்ட மலர்ச்சியான முகம், புன்னகை தவழ… நீலப்பட்டுப்புடவையின் ஒளியில் … வெள்ளிக்கம்பிகள் மின்னுவதுபோல் தலைமுடியில் இடையிடையே வெள்ளை முடி… கையில் ஒலி வாங்கி…. தம்புரா சுருதி கூட்ட ராகமாலிகாவில் குறையொன்று மில்லை மறைமூர்த்தி கண்ணா குறையொன்று மில்லை கோவிந்தா…. என்று இசைத்தார்.

வீணைக் கலைஞரான அவரின் தாயே அவருக்கு முதல் குரு. பத்து வயதில் இசைத்தட்டுக்காகப் பாடலைப் பாடிப் பதிவு செய்தவர். இசை மேதைகளின் வழிகாட்டுதல்களில் தன்னை வளர்த்துக் கொண்டவர். ஐந்தாம் வகுப்பு வரைதான் கல்வி பயில வாய்ப்புக் கிட்டியது. பதினேழு வயதில் சென்னை மியூசிக் அகாதெமியில் மேதைகள் பலர் முன்பு கச்சேரி செய்து பாராட்டைப் பெற்றவர்.

அவருக்கு மீரா திரைப்படம் மிகப்பெரிய வெற்றியைத் தந்தது. அது அவரது கடைசித் திரைப்படமாகவும் அமைந்தது. இந்தியா முழுவதிலும் உள்ள பலரின் பாராட்டுகளையும் பெற்றார். காற்றினிலே வரும் கீதம், பிருந்தாவனத்தில் கண்ணன் முதலிய பாடல்களுக்கு மிகப்பெரிய வரவேற்புக் கிடைத்தது. ஜவஹர்லால் நேரு , சரோஜினி நாயுடு போன்ற பெரியோர்களால் பாராட்டப்பட்டவர்.

ஒருமுறை காந்தியடிகளைத் தில்லியில் சந்தித்தபோது ‘இரகுபதி இராகவ இராஜாராம்’ என்ற பாடலைப் பாடினார். அவரைப் பாராட்டிய அண்ணல், மீரா எழுதிய பாடல் ஒன்றைக் குறிப்பிட்டுப் பாடச் சொன்னார். பின் சிறிது நாள்களில் முனைந்து அந்தப் பாடலைக் கற்றுப் பயிற்சி செய்தார். சென்னை வானொலி, 1947 இல் காந்தியடிகளின் பிறந்த நாளன்று அப்பாடலை ஒலிபரப்பியது. அப்பாடல் ஹரிதும் ஹரோ’ என்னும் மீரா பஜன்.

1954 இல் அவர் தாமரையணி விருது பெற்றபோது, தன்னைத் தொட்டுத் தடவிப் பாராட்டிய பார்வையிழந்த ஹெலன் கெல்லரை தன்னால் மறக்கமுடியாது என்கிறார். 1963இல் இங்கிலாந்திலும் 1966இல் ஐ.நா. அவையிலும் பாடினார். இதே ஆண்டில் அவரின் குரலில் பதிவு செய்யப்பட்ட வெங்கடேச சுப்ரபாதம் திருப்பதியில் ஒலிக்கத்தொடங்கியது.

1974 இல் நோபல் பரிசுக்கு இணையான மகசேசே விருது அவர் இசைக்குக் கிடைத்த மகுடம். இவ்விருது பெறும் முதல் இசைக்கலைஞராகவும் ஆனார். தமிழ், தெலுங்கு, கன்னடம், சமஸ்கிருதம், மலையாளம், இந்தி, மராத்தி, குஜராத்தி ஆகிய இந்திய மொழிகளிலும் ஆங்கிலத்திலும் கூடப் பாடியுள்ளார். இந்தியா, மிக உயரிய விருதான இந்திய மாமணி’ விருதளித்து அவரைச் சிறப்பித்தது.

அவருடைய பல இசைக் கக்சேரிகள் ஏதாவது ஒரு அமைப்பின் நன்கொடைக்காக நடந்தவை என்பது பெரும் மகிழ்வை அளிக்கிறது.

Question 45.
(அ) குமரிக் கடல் முனையையும் வேங்கட மலைமுகட்டையும் எல்லையாகக் கொண்ட
தென்னவர் திருநாட்டிற்குப் புகழ் தேடித்தந்த பெருமை தகைசால் தமிழன்னையைச் சாரும்.
Answer:
எழில்சேர் கன்னியாய் என்றும் திகழும் அவ்வன்னைக்கு பிள்ளைத் தமிழ் பேசி, சதகம் சமைத்து, பரணி பாடி, கலம்பகம் கண்டு, உலா தந்து, அந்தாதி கூறி, கோவை யாத்து இவற்றையெல்லாம் அணியாகப் பூட்டி, அழகூட்டி அகம்மிக மகிழ்ந்தனர் செந்நாப் புலவர்கள். இக்கருத்துகளைக் கருவாகக் கொண்டு சான்றோர் வளர்த்த தமிழ்’ என்னும் தலைப்பில் கட்டுரை எழுதுக.

சான்றோர் வளர்த்த தமிழ்

முன்னுரை:
கல் தோன்றி மண் தோன்றாக் காலத்தே முன்தோன்றி மூத்த தமிழ்’ என்னும் பழைமையுடைய செந்தமிழ் மொழியை உயர்தனிச் செம்மொழி’ என்று வரையறுத்தவர் பரிதிமாற் கலைஞர் என்று பலராலும் போற்றப்படும் வி. கோ.

சூரிய நாராயண சாஸ்திரியார் ஆவார். அகத்தியர் வளர்த்த தமிழ்’ பொதியமலைத் தமிழ் போன்ற தொடர்கள் தமிழின் பழைமையை விளக்கும் சான்றுகளாகும். உயர் தனி, செம்மை என்ற மூன்று அடைமொழிகள் கொண்டு தமிழ் விளங்கக் காரணம் என்ன என்பதைக் காண்பதே இக்கட்டுரையின் நோக்கம்.

பொருளுரை தொன்மை :
குமரிக்கண்டம் எனப்பட்ட லெமூரியாக் கண்டத்திலுள்ள மக்கள் பேசிய மொழி தமிழ் என்பது எலியர் கருத்து. மனித இனம் எப்போது தோன்றியதோ அப்போது தோன்றியது தமிழ் உலக மொழிகளுள் பழைமையும் இலக்கிய இலக்கண வளமும் உடையவை கிரேக்கம், இலத்தீன், சீனம், அரபு, சமஸ்கிருதம், தமிழ் என்பன. பிற மொழிகள் காலவெள்ளத்தில் சிதைந்து மாறுபட்டு விளங்குகின்றன. பத்தாயிரம் ஆண்டு கட்டு முன்பே பேசப்பட்டும், இன்றும் அழியாமல் நிலைத்து நிற்கும் சிறப்பான தன்மை தமிழ் மொழிக்கு அமைந்த பண்பு எனலாம்.

உயர்மொழி :
தான் பேசப்படும் நாட்டிலுள்ள பலமொழிகளுக்கும் தலைமையும், அவற்றைவிட மேன்மைத்தன்மையும் உள்ள மொழியே, உயர்மொழி எனப்படும் என்று கூறுவார் பரிதிமாற்மலைஞர். இதன்படி பார்த்தால் திராவிட மொழிகளாகிய தெலுங்கு, கன்னடம், மலையாளம், துளு ஆகிய மொழிகளுக்கு எல்லாம் தலைமையும் மேன்மையும் பெற்றிருப்பதால் தமிழ் உயர்மொழியே ஆகும்.

தமிழ் – தனிமொழி:
தான் வழங்கும் நாட்டிலுள்ள மற்றைய மொழிகளின் உதவியில்லாமல் தனித்தியங்க வல்ல ஆற்றலுடைய மொழி ‘தனிமொழி’ எனப்படும். பிறமொழிகளுக்குச் செய்யும் உதவி மிகுந்தும், பிற மொழிகள் தனக்குச் செய்யும் உதவி குறைந்தும் காணப்படுவது நம் தமிழ்மொழியில் மட்டுமே. பிற மொழிகளின் உதவி இல்லாமல் தனித்தியங்கும் ஆற்றல் பெற்றிருப்பதால் தமிழ் மொழி ‘த எனப்படும்.

தமிழ் – செம்மொழி :
‘திருந்திய பண்பும், சீர்த்த நாகரிகம் பெற்ற தூய்மொழி புகல் செம்மொழியாகும்’ என்பது செம்மொழியின் இலக்கணம். இவ்வரையறை தமிழ் மொழிக்கும் பொருந்துகிறது. தமிழ் மொழியினுள் இடர்ப்பட்ட சொல் முடிவுகளும், தெளிவற்ற பொருள் முடிவுகளும் இல்லை. சொல்லையும் சொல்லுபவன் கருதிய பொருளைக் கேட்பவன் தெளிவாக உணர முடியும் பழையன கழிதலும் புதியன புகுதலும் என்பதைத் தமிழ்மொழி இன்றளவும் ஏற்றிருப்பதால் ‘தமிழ் செம்மொழி

ஆகும். தமிழின் பொதுப்பண்பு:
மேலநாட்டு அறிஞர்களான போப், கால்டுவெல், வீரமாமுனிவர் போன்றவர் தமிழினைக்கற்று இலக்கியத்திற்கு வளம் சேர்த்தனர். மதம், மொழி, இனம், நிறம், கடந்து ‘யாதும் ஊரே யாவரும் கேளிர்’ என்று முதல் முழக்கமிட்டது தமிழ் மொழியே ஆகும். உண்பது அமிழ்தமே ஆயினும் தனியராய் உண்ணோம் என்று உணர்த்தினவர் தமிழர்.

“தீதும் நன்றும் பிறர்தர வாரா” பிறப்பொக்கும் எல்லா உயிர்க்கும்; உண்பது நாழி ; உடுப்பவை இரண்டே போன்ற உயர்ந்த சிந்தனைகளை உலகுக்கு உணர்த்தியது தமிழ் இல்வாழ்வையும், புறவாழ்வையும் அகம், புறம் என்று பிரித்து குறிஞ்சி, முல்லை மருதம், நெய்தல், பாலை என்று ஐவகை நிலம் வகுத்து, முதல், கரு, உரிப்பொருள் வகுத்து இலக்கியம் கண்டு இலக்கணம் இயம்பியது நம் செந்தமிழ் மொழியாகும்.

அரசின் கடமை :
மூவாயிரம் ஆண்டுகள் பழைமை வாய்ந்த தமிழைச் செம்மொழியாக அறிவித்தால் பிற நாட்டவர் தமிழ்மொழியைப் பயில்வர். நம் செந்தமிழ் இலக்கியம், இலக்கணம் பிற மொழியாளர்களால் ஆராயப்படும். தமிழ் உலகம் முழுவதும் ஏற்றம் பெற்று புதிய நூல்கள் ஆக்கம் பெறும். உலக அளவில் உயரிய மதிப்பு கூடும். மொழிக் களஞ்சியங்களில் தமிழ்க்கலை வெளிப்படும். முடிவுரை.

தமிழைச் செம்மொழி ஆக்கியதோடு மட்டுமன்றி வணிகத் துறையை எட்டிப்பிடித்துச் செல்வ மொழியாகவும் மாற்றிட வழிவகை செய்ய வேண்டும்.

(அல்லது)

(ஆ) குறிப்புகளைப் பயன்படுத்தி சாரணர் இயக்கம் குறித்து கட்டுரை எழுதுக. முன்னுரை – இயக்கம் – சின்னம் – பழக்கம் – பயிற்சிகளும் வழிமுறைகளும் – ஆக்கம் முடிவுரை.

முன்னுரை:
குருவியின் தலையில் பனம் பழமா! என்று எண்ணாது சின்னஞ்சிறு மீன்கள் கூட்டம் நினைத்தால் ஒரு கப்பலையே மூழ்கடித்து விடும் என்பதே சாரண இயக்கத் தத்துவமாகும். இவ்வியக்கத்தை இங்கிலாந்து நாட்டைச் சேர்ந்த பேடன் பவுல் (1857 – 1941) என்பவர் தோற்றுவித்தார்.

ஒரு சமயம் தென்னாப்பிரிக்காவுக்கும் போயருக்கும் இடையிலான போரில் லெப்டினண்ட்டாக விளங்கினார். போர்க்காலத்தில் சின்னஞ்சிறுவர்களுக்கு (12-17) பயிற்சி கொடுத்தால் முதலுதவி போன்றவை செய்யலாமே என்று எண்ணி சிந்தனையோடு நில்லாமல் செயலிலும் இறங்கினார்.

சிறுவர்களைப் பல குழுக்களாகப் பிரித்து, குழுக்களுக்குப் பறவை, விலங்கு இவற்றின் பெயர்களிட்டு போர்க்களத்தில் செல்லும் முறை, முதலுதவி செய்தல், ஒற்றாடல் போன்ற பயிற்சியளித்தார். பயிற்சிகளுக்குப்பின் சிறுவர்களின் விரைவான செயலாக்கமும், கூர்த்த மதியும் பெருமளவு பயனைத் தந்தது.

போருக்குப் பின்பு தன்னுடைய பதவியைத் துறந்து ‘சிறுவர் சாரணியம்’ என்றும் நூலை எழுதினார். 1908ல் மாணவர்களுக்கான சாரண இயக்கத்தையும், 1910ல் மாணவியருக்கான சாரணிய இயக்கத்தையும் தோற்றுவித்தார்.

இயக்கம் :
இந்தியாவில் இதனைக் கொண்டுவரும் பொருட்டு அன்னி பெசண்ட் அம்மையார் (1847 – 1933) அவர்கள் 1917ல் சாரண – சாரணிய இயக்கத்தைக் கொண்டு வந்தார். மாணவர்களுக்கு இவ்வியக்கமானது ஒற்றுமையை, கடமை தவறாமையை, உதவி செய்தலை வளர்க்கும் என உறுதியாக நம்பினார். அந்நம்பிக்கை வீண்போகாது. இன்று சாரண இயக்கமானது உயரிய அளவில் விளங்குகிறது.

சின்னம் :
ஒவ்வொரு இயக்கமும் தமக்கென ஒரு சின்னத்தோடு திகழும். சாரண இயக்கத்திற்கும் தனி சின்னம் உள்ளது. ஒவ்வொரு சாரண-சாரணியரும் அதனை அணிந்திருத்தல் அவசியமாகும். காக்கிச் சீருடையில் கழுத்தில் நீலநிற ஸ்கார்ஃப் அணிவர். ஸ்கார்ஃப்பானது கீழே விழாவண்ணம் பேட்ஜ் சொருகி வைத்திடுவர். மேலும் பெல்ட்டும், தொப்பியும், கேன்வாஸ் ஷூவும் அணிவது அவசியமாகும்.

பழக்கம் :
சாரணர் ஒருவரையொருவர் காணும் போது வலது கையால் வணக்கத்தைத் தெரிவிப்பர். அதுவும் சிறு விரலைப் பெரு விரலால் பிடித்து மூவிரலை நெற்றியில் வைத்து வணக்கம் தெரிவிப்பது வழக்கம். மேலும் கை குலுக்கும் போது இதயத்தின் பிரதிபலிப்பாய் இடது கை கொண்டே கை குலுக்க வேண்டும்.

பயிற்சிகளும் வழிமுறைகளும்:
மாணவர்களுக்கு விடுமுறை நாட்களில் பயிற்சி தரப்படுகிறது. வெளியூர்களில் முகாமிட்டு பயிற்சிகளும் அளிக்கப்படுவதுண்டு. பயிற்சி பல்வேறு விதங்களில் அமைந்திருக்கும். முதலுதவி செய்தல், கயிறு ஏறுதல், மரம் நடுதல், வளாகங்களைச் சீர்படுத்துதல், சாலை விதிகளை அறிதல் எனப் பயனுள்ள வகைகளில் பயிற்சி பெறுவர். பல்வேறு நட்புகளை முகாம்களின் போது பெறுவர்.

பயிற்சியின் முடிவில் விளையாட்டுப் போட்டிகள், கலை நிகழ்ச்சிகள் இவை நடத்தப்படும். போட்டிகளில் முதலில் வரும் மூவர்களுக்கு சான்றிதழ்களும், பரிசுகளும் வழங்கப்படுகிறது.

ஆக்கம் :
பயிற்சி முடித்த சாரணர்கள் பயிற்சியுடன் நில்லாது தொண்டுகளும் புரிவர். விழாக்களில் மக்களை வழிநடத்தும் பணியிலும், சாலைகளைச் சீரமைக்கும் பணியிலும், ஊர்களில் மரங்களை நடுதலிலும், வளாகங்களைச் சுத்தப்படுத்துதலிலும் குழுவாகச் செயல்படும் இவர்களது பணியை அடுக்கிக் கொண்டே போகலாம்.

குழுவாக இல்லாமல் வீடுகளில் தனியாக இருந்தாலும் பிறருக்கு பாம்பு, தீ, தண்ணீர் இவற்றால் துன்பம் ஏற்படும்போது முதலுதவி செய்து காப்பாற்றவும் செய்கின்றனர்.

முடிவுரை:
சாரணர் இயக்கம் சிறுவர்களுக்கான இயக்கமாக இருந்தாலும், அவர்களைச் சிகரங்களுக்கு அழைத்துச் செல்பவையாகும். ஒற்றுமையின் விளைநிலமாகத் திகழ்கின்றனர். இங்கு தூவப்படும் விதைகள் சமுதாய எழுச்சிக்கு வழிகாட்டும் என்பதில் ஐயமில்லை . வளரட்டும் சாரணர் இயக்கம். தொடரட்டும் அவர்களது பணிகள்.