Question

At 298 K the molar conductivities at infinite dilution (${\Lambda }_m^{\circ }$) of ${NH}_{4}Cl,KOH$ and $KCl$ are 152.8, 272.6 and 149.8 $S$ ${cm}^2{mol}^{-1}$ respectively. The ${\Lambda }_m^{\circ }$ of ${NH}_{4}OH$ in $S$ ${cm}^2{mol}^{-1}$ and % dissociation of 0.01 M ${NH}_{4}OH$ with ${\Lambda }_m=25.1S{cm}^2{mol}^{-1}$ at the same temperature are:

• A. 275.6, 0.91
• B. 275.6, 9.1
• C. 269.6, 9.6
• D. 30, 84

Answer 1

The correct option is B 275.6, 9.1

${\Lambda }_{m\left(N{H}_{4}OH\right)}^0={\Lambda }_{m\left(N{H}_{4}Cl\right)}^0+{\Lambda }_{m\left(KOH\right)}^0-{\Lambda }_{m\left(KCl\right)}^0$

$=152.8+272.6-149.8=275.6Sc{m}^2/mol$

The degree of dissociation

$\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{25.1}{275.6}=0.091$

Percent dissociation

$=100\times 0.091=9.1$ %