1
MCQ (Single Correct Answer)

AIPMT 2009

The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm2 and at infinite dilution is 400 mho cm2. The dissociation constant of this acid is
A
1.25 $$ \times $$ 10$$-$$6
B
6.25 $$ \times $$ 10$$-$$4
C
1.25 $$ \times $$ 10$$-$$4
D
1.25 $$ \times $$ 10$$-$$5

Explanation

The degree of dissociation ($$\alpha $$)

$$\alpha $$ = $${8 \over {400}}$$ = 2 $$ \times $$ 10-2

From Ostwald’s dilution law for weak monobasic acid, we have

kc = C$$\alpha $$2

= $${1 \over {32}} \times {\left( {2 \times {{10}^{ - 2}}} \right)^2}$$

= 1.25 $$ \times $$ 10-5
2
MCQ (Single Correct Answer)

AIPMT 2009

Given :
(i)   Cu2+ + 2e$$-$$ $$ \to $$ Cu,  Eo = 0.337 V
(ii)  Cu2+ + e$$-$$ $$ \to $$ Cu+,  Eo = 0.153 V
Electrode potential, Eo for the reaction,
Cu+ + e$$-$$ $$ \to $$ Cu,  will be
A
0.90 V
B
0.30 V
C
0.38 V
D
0.52 V

Explanation

For the reaction,

Cu2+ + 2e$$-$$ $$ \to $$ Cu,  Eo = 0.337 V

$$\Delta $$Go = - nFEo

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu2+ + e$$-$$ $$ \to $$ Cu+,  Eo = 0.153 V

$$\Delta $$Go = - nFEo

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu2+ + e$$-$$ $$ \to $$ Cu+

$$\Delta $$Go = –0.521 F = –nFE°

$$ \Rightarrow $$ E° = 0.52 V
3
MCQ (Single Correct Answer)

AIPMT 2008

Standard free energies of formation (in kJ/mol) at 298 K are $$-$$237.2, $$-$$ 394.4 and $$-$$8.2 for H2O(l), CO2(g) and pentane (g) respectively. The value of Eocell for the pentane-oxygen fuel cell is
A
1.0968 V
B
0.0968 V
C
1.968 V
D
2.0968 V

Explanation

At Anode:

C5H12 + 10H2O $$ \to $$ 5CO2 + 32H+ + 32e-

At Cathode:

8O2 + 32H+ + 32e- $$ \to $$ 16H2O
-------------------------------------------------
C5H12(g) + 8O2(g) $$ \to $$ 5CO2(g) + 6H2O(l)

$$\Delta $$G = 5×$$\Delta $$GCO2 + 6 $$\Delta $$G(H2O) – [$$\Delta $$G(C5H12) +8 × $$\Delta $$GO2 ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol–1

= – 3387 × 103 J mol–1

$$\Delta $$G = - nFEocell

From the overall equation we find n = 32

– 3387 × 103 = -32 $$ \times $$ 96500 $$ \times $$ Eocell

Eocell = $${{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}$$ = 1.0968 V
4
MCQ (Single Correct Answer)

AIPMT 2008

Kohlrausch's law states that at
A
Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
B
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C
Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
D
Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.

Explanation

Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions.

$$\lambda $$$$\infty $$ = $$\lambda $$a + $$\lambda $$c

where, $$\lambda $$a = equivalent conductance of the anion

$$\lambda $$c = equivalent conductance of the cation.

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI