Question Set A

CBSE 12th Standard Chemistry Aldehydes , Ketones and Carboxylic Acids HOT Questions Study Material - QB365

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Questions:

Generic 2

5 x 2 = 10

Rusting of iron is quicker in saline water than in ordinary water. Give reason.

Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Pentan-2-one and Pentan-3-one
(ii) Ethanal and Propanal

Compare the strength of following acids:
(i) Formic acid,
(ii) Acetic acid,
(iii) Benzoic acid.

How many Faradays of charge are required to convert:
1 mole of ${ MnO }_{ 4 }^{ - }$ to Mn²⁺ion,

What are the products obtained during electrolysis of CuSO₄using Pt electrode?

Generic 3

5 x 3 = 15

What are the signs of $\Delta G,K$ and ${ E° }_{ cell }$ for a spontaneous cell reaction?

Name the following compounds according to IUPAC system of nomenclature:
(i) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(ii) CH₃COCH₂COCH₃
(iii) (CH₃)₃CCH₂COOH

Write balanced chemical equations for the following reactions:
(i) Thionyl chloride reacts with benzoic acid
(ii) Acetic acid is reacted with red phosphorus and HI
(iii) Acetic acid is treated with Zn metal.

(a) Write the step and conditions involved in the following conversions:
(i) Acetophenone to 2-phenyl-2-butanol
(ii) Propene to acetone
(b) Describe simple chemical tests to distinguish between the following pairs of compounds:
Diethyl ether and Propanol

10.

How long will it take an electric current of 0.15 A to deposit all the copper from 500 ml of 0.15 M copper sulphate solution?

Generic 5

6 x 5 = 30

11.

(i) What is limiting molar conductivity? Why there is steep rise in the molar conductivity of a weak electrolyte on dilution?
(ii) Calculate the emf of the following cell at 298K:
Mg(s) I Mg²⁺(0.1M)|| Cu²⁺(1.0 X 10^-3 M) I Cu(s) [Given= E^ocell = 2.71 V ]

12.

Calculate the standard electrode potential of Ni²⁺ II Ni electrode if emf of the cell, Ni (s)1 Ni²⁺(0.01M) IICu²⁺(0.1 M) ICu(s) is 0.059 V. (Given $E^{ 0 }_{ cu2+/cu }=+0.34V)$

13.

Rahul visited the house of his friend Shyam and found that all the water taps were rusted. On enquiry, he came to know that these were iron taps. Rahul advised his friend to use either chrome plated or nickel plated taps. Shyam accepted his advice.
On the basis of above passage give the answer of following questions.
(i) Why did iron tap get rusted?
(ii) What was the purpose of chrome plating or nickel plating?
(iii) What is the value associated with this?

14.

An organic compound (A) on treatment with ethyl alcohol gives a carboxylic acid (B) and compound (C). Hydrolysis of (C) under acidified conditions gives (B) and (D). Oxidation of (D) with KMnO₄ also gives (B). (B) on heating with Ca(OH)2 gives (E) having moleuclar formula C₃H₆O. (E) does not give TOllens'test and does not reduce Fehiling's solution but forms 2, 4-dinitrophenyhydrazone. Identify (A),(B),(C),(D) and (E).

15.

Describe the following reactions
(i) Cannizzaro's reactions.
(ii) Cross aldol condensation

16.

(a) Identify A, B and C in the following sequence of reactions:

(b) Predict the structures of the products formed when benzaldehyde is treated with
(i) conc. NaOH
(ii) HNO₃ / H₂SO₄ (at 273 - 383 k)

Multiple Choice Question

5 x 1 = 5

17.

Two electrolytes X and Yare diluted. Λ_m of Y increases 1.5 times and for X it increases 25 times. Predict the strong electrolyte among X and Y.

(a)

(b)

(c)

Both (a) and (b)

(d)

There is no effect of dilution on nature of electrolyte.

18.

A carbonyl compound reacts with hydrogen cyanic to form a cyanhydrin which on hydrolysis forms a racemic mixture of $\alpha$ -hydroxy acid. The carbonyl compound is

(a)

formaldehyde

(b)

acetaldheyde

(c)

acetone

(d)

diethyl ketone

19.

The compound that does not undergo Cannizzaro reaction is

(a)

formaldehyde

(b)

acetaldehyde

(c)

benzaldehyde

(d)

trimethylacetaldehyde

20.

For the cell, TI | TI⁺ (0.001 M) | Cu²⁺ (0.1 M) | Cu, E_cell at 25°C is 0.83 V. This can be increased

(a)

by increasing [Cu²⁺]

(b)

by increasing [TI⁺]

(c)

by decreasing [Cu²⁺]

(d)

by decreasing [TI⁺]

21.

The positive value of the standard electrode potential of Cu²⁺ /Cu indicates that ___________________.

(a)

this redox couple is a stronger reducing agent than the H⁺ /H₂ couple.

(b)

this redox couple is a stronger oxidising agent than the H⁺ /H₂

(c)

Cu can displace H₂ from acid

(d)

Cu cannot displace H₂ from acid

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Answers Key:

Generic 2

5 x 2 = 10

In saline water, the presence of Na⁺ and Cl^- ions increases the conductance of the solution in contact with the metal surface. This accelerates the formation of Fe²⁺ion and hence that of rust, Fe₂O₃ x H₂O.

(i) Add I₂ and NaOH. P-2- one will give yellow ppt of iodoform, whereas pentan-3- one will not react.
(ii) Add I2 and NaOH. Ethanal will yellow precicipate of iodoform, whereas proponal does not react.

HCOOH > C₆H₅COOH > CH₃COOH

$\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}, \mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+} \mathrm{MnO}_4^{-}+5 e^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \text { i.e. }$ when 1 mole of ${M n O}_{4}^{-}$ $\mathrm{MnO}_4^{-} \text {changes to } \mathrm{Mn}^{2+}$ 5 Faradays of charge is required.

$\mathrm{CuSO}_4 \rightarrow \mathrm{Cu}^{2+}+\mathrm{SO}_4^{2-}$
$ \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} $
$ \mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) , At\ anode 2 \mathrm{OH}^{-} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} or 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-}$
Copper is formed at cathode and oxygen gas is liberated at anode

Generic 3

5 x 3 = 15

$\Delta G=\_ ve,K=+ve,{ E° }_{ cell }=+ve$

(i) 6-Chloro-4-ethylhexan-3-one
(ii) Pentane-2, 4-dione
(iii) 3, 3-Dimethylbutanoic acid

10.

500 ml of 0.15 M CuSO₄ solution contains $\frac { 500\times 0.15 }{ 1000 } $
Mass of Cu = 0.075 x 63.5 = 4.7625 g; Eq. wt. of Cu2⁺ =$\frac{63.5}{2}$=31.75
m = Z $\times$ l $\times$ t
4.7625 = $\frac{31.75}{96500}$$\times$0.15 $\times$ t
t= $\frac { 4.7625\times 96500 }{ 31.75\times 0.15 } $ = 96500 sec = $\frac { 96500 }{ 60\times 60 } $ = 26.80 hours.

Generic 5

6 x 5 = 30

11.

(i) When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.
The change in Λ_m with dilution is due to the increase in the degree of dissociation and consequently the number of ions in the total volume of the solution that contains 1 mol of electrolyte, hence Λ_mincreases steeply.
(ii) $E_{\text {cell }} =E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} $
$=2.71 \mathrm{~V}-\frac{0.059}{2} \log \frac{0.1}{0.001} $
$=2.71 \mathrm{~V}-\frac{0.059}{2} \log 10^{2}=2.651 \mathrm{~V} $

12.

(i) First, find E⁰_cell
(ii) Then find $E^{ 0 }_{ anode }$ by using the formula
$E^{ 0 }_{ cell }=E^{ 0 }-_{ cathode }E^{ 0 }_{ anode }$
Given E_cell = 0.059V; E⁰_cu2+/cu = +0.34V
[Ni²⁺] = 0.01M and [cu²⁺] = 0.1M
E_cell = E^₀_cell - $\frac { 0.059 }{ 2 } log\frac { \left[ Ni^{ 2+ }(aq) \right] }{ \left[ Cu^{ 2+ }(aq) \right] } $
$0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { \left( 0.01 \right) }{ 0.1 } \left( \because n=2 \right) $
$0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { 1 }{ 10 } \left[ \because log10^{ -1 }=-1 \right] $
$\because 0.059=E^{ 0 }_{ cell }+0.0295 \times 1$
$E^{ 0 }_{ cell }=0.059-0.0295$
= 0.0295 V = 0.03 V
$E^{ 0 }_{ cell }=E^{ 0 }_{ cathode }E^{ 0 }_{ anode }$
0.03 = 0.34 - E⁰_anode

or$E^{ 0 }_{ anode }=E^{ 0 }_{ Ni2+/Ni }=0.34-0.03$
_{= 0.31 V}

13.

(i) Iron is p e to rust. It gets rusted when kept in open because it gets oxides in presence air and moisture.
(ii) The purpose of depositing a layer of chromium or nickel on the surface of iron is to check rusting. These metals are not affected by air or moisture.
(iii) (a) General awareness
(b) Friendship
(c) Skill of applying knowledge of chemistry

14.

(i) Since compound (E) with molecular formula, C₃H₆0 does not reduce Tollens' reagent and Fehling's solution but forms 2, 4-dinitrophenylhydrazone, it must be a ketone. But the only possible ketone having the molecular formula, C₃H₆O is acetone or propanone. Thus, compound (E) is acetone or (propanone) CH₃COCH3·
(ii) Since acetone (E) is obtained by heating compound (8) with Ca(OH)₂ therefore, (B) must be acetic acid (ethanoic acid), CH₃COOH.
(iii) Since (D) on oxidation with KMn04 gives acetic acid (8), therefore, (D) must be ethyl alcohol (ethanol), CH₃CH₂OH.
(iv) Since acetic acid (8) and ethyl alcohol (D) are obtained by hydrolysis of (C) under acidic conditions, therefore, (C) must be ethyl acetate (ethyl ethanoate), CH₃COOC₂H₅
(v) Since ethyl acetate (C) and acetic acid (8) are obtained by treatment of compound (A) with ethyl alcohol, therefore, compound (A) must be acetic anhydride (ethanoic anhydride), (CH₃COO)₂O.
(vi) All the reactions involved in this problem can now be explained as follows