1

### 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.
2

### AIPMT 2008

On the basis of the following Eo values, the strongest oxidizing agent is
[Fe(CN)6]4$-$ $\to$ [Fe(CN)6]3$-$ + e$-$;  Eo = $-$0.35 V

Fe2+ $\to$ Fe3+ + e$-$;  Eo = $-$0.77 V
A
Fe3+
B
[Fe(CN)6]3$-$
C
[Fe(CN)6]4$-$
D
Fe2+

## Explanation

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN)6]4$-$ $\to$ [Fe(CN)6]3$-$ + e$-$;  Eo = $-$0.35 V

Fe2+ $\to$ Fe3+ + e$-$;  Eo = $-$0.77 V

Higher the +ve reduction potential, stronger will be the oxidising agent. Oxidising agent oxidises other compounds and gets itself reduced easily.
3

### AIPMT 2007

The efficiency of a fuel cell is given by
A
$\Delta$G/$\Delta$S
B
$\Delta$G/$\Delta$H
C
$\Delta$S/$\Delta$G
D
$\Delta$H/$\Delta$G

## Explanation

Efficiency of a fuel cell ($\phi$) = ${{\Delta G} \over {\Delta H}} \times 100$

Generally, fuel cells are expected to have an efficiency of 100 percent.
4

### AIPMT 2007

The equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) $\to$ Cu2+(aq) + 2Ag(s);
Eo = 0.46 V at 298 K is
A
2.0 $\times$ 1010
B
4.0 $\times$ 1010
C
4.0 $\times$ 1015
D
2.4 $\times$ 1010

## Explanation

RT ln K = nFE°

ln K = ${{nFE^\circ } \over {RT}}$

= ${{2 \times 0.46} \over {0.0591}}$

$\Rightarrow$ K = 4 $\times$ 1015