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JEE MAIN - Chemistry (2025 - 4th April Evening Shift - No. 24)

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, its degree of dissociation is given by $x \times 10^{-1}$. The value of $x$ is __________ . (Nearest integer)
Answer
3

Explanation

$$\begin{aligned} & \lambda_{\mathrm{m}}^{\circ} \text { of } \mathrm{NH}_4 \mathrm{Cl}=185 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}=185 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}=185-70=115 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{OH}^{-}} \\ & =115+170 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=285 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \end{aligned}$$

$$\begin{aligned} \text { degree of dissociation } & =\frac{\left(\lambda_{\mathrm{m}}\right)_{\mathrm{NH}_4 \mathrm{OH}}}{\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}} \\ & =\frac{85.5}{285} \\ & =0.3 \\ & =3 \times 10^{-1} \end{aligned}$$

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