1
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
2
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
3
MCQ (Single Correct Answer)

AIPMT 2012 Mains

Standard reduction potentials of the half reactions are given below :
F2(g) + 2e$$-$$ $$ \to $$ 2F$$-$$(aq) ;   Eo = + 2.85 V
Cl2(g) + 2e$$-$$ $$ \to $$ 2Cl$$-$$(aq) ;   Eo = + 1.36 V
Br2(l) + 2e$$-$$ $$ \to $$ 2Br$$-$$(aq) ;   Eo = + 1.06 V
I2(s) + 2e$$-$$ $$ \to $$ 2I$$-$$(aq) ;  Eo = + 0.53 V

The strongest oxidising and reducing agents 23 respectively are
A
F2 and I$$-$$
B
Br2 and Cl$$-$$
C
Cl2 and Br$$-$$
D
Cl2 and I2

Explanation

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.
4
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

Limiting molar conductivity of NH4OH
$$\left[ {} \right.$$i.e.  $$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$$$\left. {} \right]$$ is equal to
A
$$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {NaOH} \right)}^0$$
B
$$\Lambda _{m\left( {NaOH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {N{H_4}Cl} \right)}^0$$
C
$$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0$$
D
$$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$

Explanation

According to Kohlrausch’s law, the molar conductivity of NH4OH

$$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$ = $$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$

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