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

Chemistry · Electrochemistry

JEE Main 2025 — 4 April, Evening Shift — Question 17

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

Numerical answer — enter this value.

Step-by-step solution

$\wedge_{\mathrm{m}}^{0}\left(\mathrm{NH}_{4} \mathrm{Cl}\right)=185 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ ,

$\lambda_{\text {eq. }}\left(\mathrm{OH}^{-}\right)=170 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ ,

$\lambda_{\text {eq. }}\left(\mathrm{Cl}^{-}\right)=70 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ ,

$\wedge^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=\lambda^{0}\left(\mathrm{NH}_{4}^{+}\right)+\lambda^{0}\left(\mathrm{OH}^{-}\right)$

$=\wedge^{0}\left(\mathrm{NH}_{4} \mathrm{Cl}\right)-\lambda^{0}\left(\mathrm{Cl}^{-}\right)+\lambda^{0}\left(\mathrm{OH}^{-}\right)$

$=(185-70)+170$

$=285 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

$\alpha=\frac{85.5}{285}=0.3=3 \times 10^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Conductance of Solutions and Kohlrausch's Law