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JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2025 · 28 January · Morning Shift · Chemistry · Electrochemistry

Given below is the plot of the molar conductivity vs concentration...

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)

Correct answer

0

Step-by-step solution

\begin{aligned} & \mathrm{R}=\rho \frac{\ell}{\mathrm{A}} \\ & \kappa=\mathrm{G} \cdot \mathrm{G}^* \quad \mathrm{G}=\frac{1}{\mathrm{R}} ; \kappa=\frac{1}{\rho}\end{aligned}

\begin{aligned} \mathrm{G}^*= & \frac{\ell}{\mathrm{A}} \\ \mathrm{R} & =\text{ Resistance } \\ \rho & =\text{ Resistivity } \\ \frac{\ell}{\mathrm{A}} & =\text{ cell constant }\left(\mathrm{G}^*\right)\end{aligned}

\begin{aligned} & \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_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\left(\lambda_{\mathrm{m}} \cdot \mathrm{C}\right)}{\left(\lambda_{\mathrm{m}} \cdot C\right)_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{R_{\mathrm{c}}} \quad \\ & \mathrm{c}=\text{ concentrated sol. } \quad \mathrm{d} \text{ = diluted solution } \\ & \frac{100 \cdot(0.15)^2}{150 .(0.1)^2}=\frac{R_{\mathrm{d}}}{100} \\ & \mathrm{R}_{\mathrm{d}}=150 \Omega\end{aligned}

$

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