Question

The equivalent conductance of an infinitely dilute solution of $N{H}_{4}Cl$ is $150$ and the ionic conductance of $O{H}^{–}$ and $C{l}^{–}$ ions are $198$ and $76$ respectively. If the equivalent conductance of a $0.01N$ solution of $N{H}_{4}OH$ is $9.6$, what will be its degree of dissociation?

• A. $0.96$
• B. $0.0103$
• C. $0.414$
• D. $0.0353$

Answer 1

The correct option is D $0.0353$

${\Lambda }_{eq}^{\circ }\left(N{H}_{4}Cl\right)={\Lambda }_{eq}^{\circ }\left(N{H}_4^{+}\right)+{\Lambda }_{eq}^{\circ }\left(C{l}^{-}\right)$
${\Lambda }_{eq}^{\circ }\left(N{H}_{4}Cl\right)=150={\Lambda }_{eq}^{\circ }\left(N{H}_4^{+}\right)+76$
${\Lambda }_{eq}^{\circ }\left(N{H}_4^{+}\right)=150-76=74$
${\Lambda }_{eq}^{\circ }\left(N{H}_{4}OH\right)={\Lambda }_{eq}^{\circ }\left(N{H}_4^{+}\right)+{\Lambda }_{eq}^{\circ }\left(O{H}^{-}\right)$
${\Lambda }_{eq}^{\circ }\left(N{H}_{4}OH\right)=74+198=272$
As, ${\Lambda }_{eq}\left(N{H}_{4}OH\right)=9.6$ (given)
So, degree of dissociation will be:
$\alpha =\frac{{\Lambda }_{eq}\left(N{H}_{4}OH\right)}{{\Lambda }_{eq}^{\circ }\left(N{H}_{4}OH\right)}=\frac{9.6}{272}=0.0353$
Hence, option (a) is correct answer.