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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2026 · 28 January · Evening Shift · Chemistry · Electrochemistry

For strong electrolyte _ m increases slowly with dilution and can...

For strong electrolyte Λm\Lambda_{\mathrm{m}} increases slowly with dilution and can be represented by the equation \Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}^{\circ}-Ac1/2\mathrm{Ac}^{1 / 2}MolarconductivityvaluesofthesolutionsofstrongelectrolyteABatMolar conductivity values of the solutions of strong electrolyte AB at18^{\circ} C\mathrm{C} are given below : (c  [mol L−1]0.040.090.160.25Λm  [S cm2 mol−1]96.195.795.394.9\begin{array}{|l|c|c|c|c|} \hline \mathrm{c}\;[\mathrm{mol\,L^{-1}}] & 0.04 & 0.09 & 0.16 & 0.25 \\ \hline \Lambda_{\mathrm{m}}\;[\mathrm{S\,cm^{2}\,mol^{-1}}] & 96.1 & 95.7 & 95.3 & 94.9 \\ \hline \end{array}) The value of constant A based on the above data [\left[\right. in Scm2mol−1/(mol/L)1/2]\left.\mathrm{S} \mathrm{cm}^{2} \mathrm{ mol}^{-1} /(\mathrm{mol} / \mathrm{L})^{1 / 2}\right] unit is ____\_\_\_\_.

Correct answer

0

Step-by-step solution

The equation Λm=Λm0−Ac1/2\Lambda_m = \Lambda_m^0 - Ac^{1/2} is linear in c1/2c^{1/2}.
Taking two data points: (c=0.04, c\sqrt{c}=0.2, Λm\Lambda_m=96.1) and (c=0.25, c\sqrt{c}=0.5, Λm\Lambda_m=94.9).
Slope = 94.9−96.10.5−0.2=−1.20.3=−4.0\frac{94.9 - 96.1}{0.5 - 0.2} = \frac{-1.2}{0.3} = -4.0
Since slope = -A, the constant A = 4 S cm² mol⁻¹/(mol/L)^(1/2)
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