1

### AIPMT 2005

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is
(Atomic mass : Al = 27)
A
270 kg
B
540 kg.
C
90 kg
D
180 kg

## Explanation

2Al2O3 + 3C $\to$ 4Al + 3CO2

From the above equation,

3 mol × 12 g mol–1 = 36 g of carbon is consumed

to give 4 mol × 27 g mol–1 = 108 g of Al

So, 108 g of Al is produced by 36 g of carbon

$\therefore$ 270000 g of Al is produced by

= ${{36} \over {108}} \times 270000$ of C

= 90000 g of C

= 90 kg of C
2

### AIPMT 2005

4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H ions in solution by the same quantity of electric charge will be
A
44.8 L
B
22.4 L
C
11.2 L
D
5.6 L

## Explanation

${{{M_{Al}}} \over {{M_H}}} = {{{E_{Al}}} \over {{E_H}}}$

$\Rightarrow$ ${{4.5} \over {{M_H}}} = {{{{27} \over 3}} \over 1}$

$\Rightarrow$MH = 0.5 g

We know, Volume of 2 g H2 at STP = 22.4 L

$\therefore$ Volume of 0.5 g H2 at STP

= ${{22.4 \times 0.5} \over 2}$ = 5.6 L
3

### AIPMT 2004

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25oC. The equilibrium constant of the reaction would be
A
2.0 $\times$ 1011
B
4.0 $\times$ 1012
C
1.0 $\times$ 102
D
1.0 $\times$ 1010

## Explanation

We know, from Nernst Equation

Ecell = Eocell - ${{2.303RT} \over {nF}}{\log _{10}}K$

At equilibrium Ecell = 0

$\therefore$ 0 = Eocell - ${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$

$\Rightarrow$ 0 = 0.295 - ${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$

$\Rightarrow$ 0.295 = ${{0.0591} \over 2}{\log _{10}}K$

$\Rightarrow$ ${\log _{10}}K$ = 10

$\Rightarrow$ K = 1 $\times$ 1010
4

### AIPMT 2003

On the basis of the information available from the reaction,

4/3Al + O2 $\to$ 2/3Al2O3,   $\Delta$G = $-$ 827 kJ mol$-$1 of O2,

the minimum e.m.f. required to carry out an electrolysis of Al2O3 is
(F = 96500 C mol$-$1)
A
2.14 V
B
4.28 V
C
6.42 V
D
8.56 V

## Explanation

4/3Al + O2 $\to$ 2/3Al2O3,   $\Delta$G = $-$ 827 kJ mol$-$1

For 1 mol of Al, n = 3

$\therefore$ For ${4 \over 3}$ mol of Al, n = $3 \times {4 \over 3} = 4$

As $\Delta$G = - nFEo

$\Rightarrow$ – 827 × 103 J = – 4 × E° × 96500

$\Rightarrow$ Eo = ${{827 \times {{10}^3}} \over {4 \times 96500}}$ = 2.14 V

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