JEE Advanced 2021 — Paper 2 — Question 34

$\alpha_{1}=\frac{\Lambda_{\mathrm{m}}^{\mathrm{c}}}{\Lambda_{\mathrm{m}}^{0}}=\frac{\mathrm{y} \times 10^{2}}{4 \times 10^{2}}=\frac{\mathrm{y}}{4}=\alpha$ On dilution conductivity increases three times $\alpha_{2}=\frac{3 \mathrm{y} \times 10^{2}}{4 \times 10^{2}}=3 \alpha_{1}=3 \alpha$ $K_{a}=\frac{C \alpha^{2}}{1-\alpha}$ Since temperature is constant $\mathrm{K}_{\mathrm{a}}$ will be constant $\frac{\mathrm{C}_{1} \alpha_{1}^{2}}{1-\alpha_{1}}=\frac{\mathrm{C}_{2} \alpha_{2}^{2}}{1-\alpha_{2}}$ $\frac{\mathrm{C} \times \alpha^{2}}{1-\alpha}=\frac{\left(\frac{\mathrm{C}}{20}\right)(3 \alpha)^{2}}{1-3 \alpha}$ $\frac{1}{1-\alpha}=\frac{9}{20}-\frac{1}{(1-3 \alpha)}$ $20-60 \alpha=9-9 \alpha$; $\alpha=\frac{11}{51}=0.2156$ $\alpha=0.22$