1

### NEET 2013

At 25oC molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm$-$1 cm2 mol$-$1 and at infinite dilution its molar conductance is 238 ohm$-$1 cm2 mol$-$1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
A
4.008%
B
40.800%
C
2.080%
D
20.800%

## Explanation

Degree of dissociation($\alpha$) = ${{9.54} \over {238}}$ = 0.04008 = 4.008%
2

### NEET 2013

A button cell used in watches function as following.
Zn(s) + Ag2O(s) + H2O(l) $\rightleftharpoons$ 2Ag(s) + Zn2+(aq) + 2OH$-$(aq)

If half cell potentials are
Zn2+(aq) + 2e$-$ $\to$ Zn(s);  Eo = $-$0.76 V
Ag2O(s) + H2O(l) + 2e$-$ $\to$ 2Ag(s) + 2OH$-$(aq), Eo = 0.34 V

The cell potential will be
A
0.84 V
B
1.34 V
C
1.10 V
D
0.42 V

## Explanation

At anode :Zn2+(aq) + 2e$-$ $\to$ Zn(s);  Eo = $-$0.76 V

At cathode : Ag2O(s) + H2O(l) + 2e$-$ $\to$ 2Ag(s) + 2OH$-$(aq), Eo = 0.34 V

cell = E°cathode – E°anode

= 0.34 – (–0.76) = 1.10 V
3

### AIPMT 2012 Mains

Molar conductivities $\left( {\Lambda _m^o} \right)$ at infinite dilution of NaCl, Hcl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol$-$1 respectively. $\left( {\Lambda _m^o} \right)$ for CH3COOH will be
A
425.5 S cm2 mol$-$1
B
180.5 S cm2 mol$-$1
C
290.8 S cm2 mol$-$1
D
390.5 S cm2 mol$-$1

## Explanation

$\Lambda$oCH3COOH = $\Lambda$oCH3COONa + $\Lambda$oHCl - $\Lambda$oNaCl

= = 91 + 425.9 – 126.4 = 390.5
4

### AIPMT 2012 Mains

The Gibb's energy for the decomposition of Al2O3 at 500oC is as follows
${2 \over 3}$ Al2O3 $\to$ ${4 \over 3}$ Al + O2
$\Delta$rG = +960 kJ mol$-$1
The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500oC is at least
A
4.5 V
B
3.0 V
C
2.5 V
D
5.0 V

## Explanation

We know,

$\Delta$Go = – nFEo

${2 \over 3}$ Al2O3 $\to$ ${4 \over 3}$ Al + O2

Total number of Al atoms in Al2O3

= ${2 \over 3} \times 2 = {4 \over 3}$

Al3+ + 3e $\to$ Al

As 3e change occur for each Al atom

$\therefore$ n = ${4 \over 3} \times 3 = 4$

Eo = - ${{\Delta G^\circ } \over {nF}}$

= - ${{960 \times 1000} \over {4 \times 96500}}$

= - 2.5 V