1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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