Resistance of a conductivity cell (cell… - JEE Main 2022 Chemistry PYQ

JEE Main 2022 — Chemistry Physical Chemistry

2022integerhard

Resistance of a conductivity cell (cell constant $129{m}^{-1}$ ) filled with $74.5\mathrm{ppm}$ solution of $\mathrm{KCl}$ is $100\Omega$ (labelled as solution $1$ ). When the same cell is filled with $\mathrm{KCl}$ solution of $149\mathrm{ppm}$ , the resistance is $50\Omega$ (labelled as solution $2$ ). The ratio of molar conductivity of solution $1$ and solution $2$ is i.e. $\frac{{\wedge }_{1}}{{\wedge }_{2}}=x\times {10}^{-3}$ . The value of $x$ is____Given, molar mass of $\mathrm{KCl}$ is $74.5g{\mathrm{mol}}^{-1}$ )

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Held on 29 Jul 2022 · Verified 6 Jul 2026.

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Solution

Given $\frac{l}{A}=129{m}^{-1}$

$\mathrm{KCl}$ solution $1$ :

$74.5\mathrm{ppm},{R}_{1}=100\Omega$

$\mathrm{KCl}$ solution $2$ :

$149\mathrm{ppm},{R}_{2}=50\Omega$

Here, $\frac{{\mathrm{ppm}}_{1}}{{\mathrm{ppm}}_{2}}=\frac{{M}_{1}}{{M}_{2}}$

and $\kappa =\frac{1}{R}G*\mathrm{where}G*=\mathrm{cell}\mathrm{constant}$

Since value of cell constant is same in both the cases.

$\frac{{k}_{1}}{{k}_{2}}=\frac{{R}_{2}}{{R}_{1}}$

$\frac{{\wedge }_{1}}{{\wedge }_{2}}=\frac{{k}_{1}\times \frac{1000}{{M}_{1}}}{{k}_{2}\times \frac{1000}{{M}_{2}}}$

$=\frac{{K}_{1}}{{K}_{2}}\times \frac{{M}_{2}}{{M}_{1}}=\frac{{R}_{2}}{{R}_{1}}\times \frac{{M}_{2}}{{M}_{1}}$

$=\frac{50}{100}\times \frac{149}{74.5}=1$

$=\frac{{\wedge }_{1}}{{\wedge }_{2}}=1,000\times {10}^{-3}$

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