If the molar conductivity ( Lambda m ) of a 0.050 mol L -1 solution of a monobasic weak acid is 90 S cm 2 mol -1 , its extent (degree) of dissociation will be [Assume Lambda + ° =349.6 S cm 2 mol -1 and Lambda - ° =50.4 S cm 2 mol -1 .]

Chemistry · Electrochemistry

NEET (UG) 2025 — Question 56

If the molar conductivity ( $\Lambda_{\mathrm{m}}$ ) of a $0.050 \mathrm{~mol} \mathrm{~L}^{-1}$ solution of a monobasic weak acid is $90 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ , its extent (degree) of dissociation will be [Assume $\Lambda_{+}^{\circ}=349.6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $\Lambda_{-}^{\circ}=50.4 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ .]

Option A:
0.115
Option B:
0.125
Option C:
0.225
Correct
Option D:
0.215

Answer: C

Step-by-step solution

Degree of dissociation $(\alpha)=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}}$

\begin{aligned} \hat{\mathrm{m}}_{\mathrm{m}}^{\circ} & =\Lambda_{+}^{\circ}+\Lambda_{-}^{\circ} \\& =349.6+50.4=400 \mathrm{Scm}^{2} \mathrm{~mol}^{-1} \\& \therefore \alpha =\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{90}{400}=0.225 \end{aligned}

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Chemistry
Chapter: Electrochemistry
Topic: Conductance of Solutions and Kohlrausch's Law