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1

### AIPMT 2009

MCQ (Single Correct Answer)
Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 $$\times$$ 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is product? (Assume 100% current efficiency, at mass of Al = 27 g mol$$-$$1).
A
8.1 $$\times$$ 104 g
B
2.4 $$\times$$ 105 g
C
1.3 $$\times$$ 104 g
D
9.0 $$\times$$ 103 g

## Explanation

E = Z × 96500

$$\Rightarrow$$ $${{27} \over 3}$$ = Z $$\times$$ 96500

$$\Rightarrow$$ Z = $${9 \over {96500}}$$

Now applying the formula, W = Z × I × t

W = $${9 \over {96500}}$$ $$\times$$ 4 $$\times$$ 104 $$\times$$ 6 $$\times$$ 60 $$\times$$ 60

= 8.1 × 104 g
2

### AIPMT 2009

MCQ (Single Correct Answer)
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
3

### AIPMT 2009

MCQ (Single Correct Answer)
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
4

### AIPMT 2008

MCQ (Single Correct Answer)
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

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