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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 4 April · Evening Shift · Chemistry · Electrochemistry

The molar conductance of an infinitely dilute solution of ammonium...

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm2 mol−1185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm2 mol−70 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-} 1{ }^1, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm2 mol−185.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, its degree of dissociation is given by x×10−1\mathrm{x} \times 10^{-1}. The value of xx is ________ (Nearest integer)

Correct answer

0

Step-by-step solution

\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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