QJEEAdv21Chf289: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Chf289 Solution: At 298 K, the limiting molar conductivity of a weak monobasic…

Degree of dissociation = $\alpha$

Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 10² S cm² mol ^{$-$ 1}

Molar conductivity, $\Lambda _m$ = y $\times$ 10² S cm² mol ^{$-$ 1}

Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 10² S cm² mol ^{$-$ 1}

Concentration before dilution = C

Concentration after dilution = ${C \over {20}}$

Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)

Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$

Putting Eq. (i),

${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$

Dissociation constant before dillution,

${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)

Dissociation constant after dilution,

${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)

Comparing Eqs. (ii) and (iii),

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

${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$

${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$

$80 - 60y = 36 - 9y$

$80 - 36 = 60y - 9y$

$44 = 51y$

$\Rightarrow y = {{44} \over {51}}$ .... (iv)

Putting in Eq. (i),

$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{y \times {{10}^2}} \over {4 \times {{10}^2}}}$

$\alpha = {{44} \over {51 \times 4}} = {{11} \over {51}}$

$\alpha = 0.22$

The value of $\alpha$ is 0.22.