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JEE Advanced 2021 Paper 2

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298\,\mathrm{K}

, the limiting molar conductivity of a weak monobasic acid is

4 \times 10^{2}\,\mathrm{S}\,\mathrm{cm}^{2}\,\mathrm{mol}^{-1}

. At

298\,\mathrm{K}

, for an aqueous solution of the acid the degree of dissociation of

\alpha

and the molar conductivity is

\boldsymbol{y} \times 10^{2}\,\mathrm{S}\,\mathrm{cm}^{2}\,\mathrm{mol}^{-1}

. At

298\,\mathrm{K}

, upon 20 times dilution with water, the molar conductivity of the solution becomes

3y \times 10^{2}\,\mathrm{S}\,\mathrm{cm}^{2}\,\mathrm{mol}^{-1}

Section: Chemistry

Question : 26 of 57

Marks: +1, -0

The value of

\alpha

Solution:

K_{a} = \; \frac{\Lambda_{m}^{2} C}{\Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m})}

K_{a} = \; \frac{(y \times 10^{2})^{2} \times C}{4 \times 10^{2}(4 \times 10^{2} - y \times 10^{2})}

= \; \frac{(3y \times 10^{2}) \times \frac{C}{20}}{4 \times 10^{2}(4 \times 10^{2} - 3y \times 10^{2})}

\Rightarrow \; \frac{1}{4-y} = \; \frac{9}{20(4-3y)} \Rightarrow y = \; \frac{44}{51}

\alpha = \; \frac{\; \frac{44}{51} \times 10}{4 \times 10^{2}}

\alpha = 0.2156(\alpha = 0.22

0.21)

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