At $25^{\circ} \mathrm{C}$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9.54…
- $2.080 \%$
- $20.800 \%$
- $4.008 \%$
- $40.800 \%$
Solution
$\lambda_c=9.54 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Molar conductance at infinite dilution,
$\lambda_c^{\infty}=238 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
We know that, degree of ionisation,
$\begin{aligned} \alpha & =\frac{\lambda_c}{\lambda_c^{\infty}} \times 100 \\ & =\frac{9.54}{238} \times 100=4.008 \%\end{aligned}$
Asked in: NEET 2013 (All India)