JEE Advanced 2021 Paper 2 · Q07 · Electrolysis & Faraday's Laws

Correct answer:0.22

Solution

Degree of dissociation $\alpha = \Lambda_m/\Lambda_m^\circ = (y \times 10^2)/(4 \times 10^2) = y/4$ $\Rightarrow$ $\alpha_1 = y/4$.

After 20-fold dilution, concentration $c_2 = c_1/20$, molar conductivity $= 3y \times 10^2 \Rightarrow$ $\alpha_2 = (3y\times10^2)/(4\times10^2) = 3y/4 = 3\alpha_1$.

For a weak monobasic acid, $K_a = c\alpha^2/(1-\alpha)$ is constant at constant temperature: $\frac{c_1\alpha_1^2}{1-\alpha_1} = \frac{c_2\alpha_2^2}{1-\alpha_2} = \frac{(c_1/20)(3\alpha_1)^2}{1-3\alpha_1}$ $\Rightarrow \frac{\alpha_1}{1-\alpha_1} = \frac{9\alpha_1/20}{1-3\alpha_1}$ $\Rightarrow 20(1-3\alpha_1) = 9(1-\alpha_1)$ $\Rightarrow 20 - 60\alpha_1 = 9 - 9\alpha_1$ $\Rightarrow 11 = 51\alpha_1$ $\Rightarrow \alpha_1 = 11/51 \approx 0.2157 \approx 0.22$.

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