Question

At $298K$ the molar conductivities at infinite dilution $\left(\Lambda m^{\circ }\right)$ of $N{H}_{4}Cl,KOH$ and $KCl$ are $152.8$ , $272.6$ and $149.8Sc{m}^{2}mo{l}^{-1}$ respectively. The $\left(\Lambda m^{\circ }\right)$ of $N{H}_{4}OH$ in $Sc{m}^{2}mo{l}^{-1}$ and dissociation of $0.01MN{H}_{4}OH$ with ${\Lambda }_m=25.1Sc{m}^{2}mo{l}^{-1}$ at the same temperature are

• A. 275.6, 0.91
• B. 275.6, 9.1
• C. 266.6, 9.6
• D. 30, 84

Answer 1

Answer: (B)

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

$=152.8+272.6-149.8$

$=275.6Sc{m}^{2}mo{l}^{-1}$

Degree of dissociation $(\alpha )=\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{25.1}{275.6}$

$=0.091$

$\therefore \%$ degree of dissociation $=9.1\%$