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Chemistry · Electrochemistry

JEE Main 2026 — 28 January, Evening Shift — Question 67

For strong electrolyte Λm\Lambda_{\mathrm{m}} increases slowly with dilution and can be represented by the equation Λm=Λ∘m−Ac1/2\Lambda_{\mathrm{m}}=\Lambda^{\circ}{ }_{\mathrm{m}}-\mathrm{Ac}^{1 / 2} Molar conductivity values of the solutions of strong electrolyte AB at 18∘C18^{\circ} \mathrm{C} are given below :

c[molL−1]\mathrm{c}\left[\mathrm{mol} \mathrm{L}^{-1}\right]0.040.090.160.25
Λm[Scm2 mol−1]\Lambda_{\mathrm{m}}\left[\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1}\right]96.195.795.394.9

The value of constant A based on the above data [ in Scm2 mol−1/(mol/L)1/2\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} /(\mathrm{mol} / \mathrm{L})^{1 / 2} ] unit is ____\_\_\_\_ .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

Using equation : Λm=Λm0−Ac\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{0}-\mathrm{A} \sqrt{\mathrm{c}} 96.1=Λ0m−A0.0496.1=\Lambda^{0}{ }_{\mathrm{m}}-\mathrm{A} \sqrt{0.04}

96.1=\Lambda^{0}{ }_{\mathrm{m}}-\mathrm{A} \times 0.2 \end{gathered}$$ $95.7=\Lambda^{0}{ }_{\mathrm{m}}-\mathrm{A} \times \sqrt{0.09}$ $$\begin{gathered} 95.7=\Lambda^{0}{ }_{\mathrm{m}}-\mathrm{A} \times 0.3 \end{gathered}$$ From eq. (1) and eq. (2) $\mathrm{A}=4$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law

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