Chemistry · Electrochemistry

JEE Main 2025 — 28 January, Morning Shift — Question 45

Given below is the plot of the molar conductivity vs $\sqrt{\text { concentration }}$ for KCl in aqueous solution.

If, for the higher concentration of KCl solution, the resistance of the conductivity cell is $100 \Omega$ , then the resistance of the same cell with the dilute solution is ' $x$ ' $\Omega$ . The value of $x$ is ___ (Nearest integer)

Answer: 150

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{R}=\rho \frac{\ell}{\mathrm{A}}$

$\kappa=\mathrm{G} \cdot \mathrm{G}^{*} \quad \mathrm{G}=\frac{1}{\mathrm{R}} ; \kappa=\frac{1}{\rho}$ $\mathrm{G}^{*}=\frac{\ell}{\mathrm{A}}$

$\mathrm{R}=$ Resistance $\rho=$ Resistivity $\frac{\ell}{\mathrm{A}}=$ cell constant $\left(\mathrm{G}^{*}\right)$

$\frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{\mathrm{R}_{\mathrm{c}}} ; \lambda_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathrm{C}}$

$\frac{\kappa_{c}}{\kappa_{d}}=\frac{\left(\lambda_{\mathrm{m}} \cdot C\right)}{\left(\lambda_{\mathrm{m}} \cdot C\right)_{d}}=\frac{R_{d}}{R_{c}} \quad \begin{aligned} & \mathrm{c}=\text { concentrated sol. } . d=\text { diluted solution }\end{aligned}$

$\frac{100 .(0.15)^{2}}{150 .(0.1)^{2}}=\frac{\mathrm{R}_{\mathrm{d}}}{100}$ $\mathrm{R}_{\mathrm{d}}=150 \Omega$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Conductance of Solutions and Kohlrausch's Law