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1

### AIPMT 2010 Prelims

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
A
increase in ionic mobility of ions
B
100% ionisation of electrolyte at normal dilution
C
increase in both i.e., number of ions and ionic mobility of ions
D
increase in number of ions.

## Explanation

Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions increases which in turn increases equivalent conductance of the solution.
2

### AIPMT 2009

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
3

### 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
4

### 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

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