JEE MAIN - Chemistry (2025 - 28th January Morning Shift - No. 24)
Given below is the plot of the molar conductivity vs $\sqrt{\text { concentration }}$ for KCl in aqueous solution.
_28th_January_Morning_Shift_en_24_1.png)
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
Explanation
$$\begin{aligned}
& \mathrm{R}=\rho \frac{\ell}{\mathrm{A}} \\
& \kappa=G \cdot G^* \quad G=\frac{1}{\mathrm{R}} ; \kappa=\frac{1}{\rho} \\
& \mathrm{G}^*=\frac{\ell}{\mathrm{A}} \\
& \mathrm{R}=\text { Resistance } \\
& \rho=\text { Resistivity } \\
& \frac{\ell}{\mathrm{A}}=\text { cell } \operatorname{constant}\left(\mathrm{G}^*\right) \\
& \frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{\mathrm{R}_{\mathrm{c}}} ; \lambda_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathrm{C}} \\
& \frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\left(\lambda_{\mathrm{m}} \cdot \mathrm{C}\right)}{\left(\lambda_{\mathrm{m}} \cdot C\right)_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{R_{\mathrm{c}}} \quad \begin{array}{l}
\mathrm{c}=\text { concentrated sol. } . \\
\mathrm{d}=\text { diluted solution }
\end{array} \\
& \frac{100 \cdot(0.15)^2}{150 .(0.1)^2}=\frac{\mathrm{R}_{\mathrm{d}}}{100} \\
& \mathrm{R}_{\mathrm{d}}=150 \Omega
\end{aligned}$$
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