At 298 ~K , the limiting molar conductivity of a weak monobasic...

$\alpha = {\lambda}_{m}{\lambda}_{m}^{∘}$

${\alpha}_{1} = \dfrac{y \times {10}^{+ 2}}{4 \times {10}^{2}} = \dfrac{y}{4} . . . \left(1\right)$

${k}_{a} = \dfrac{{C}_{1}{\alpha}_{1}^{2}}{1 - {\alpha}_{1}} . . . \left(2\right)$

After dilution

${\alpha}_{2} = \dfrac{3y \times {10}^{2}}{4 \times {10}^{2}} = \dfrac{3y}{4} . . . \left(3\right)$

${k}_{a} = \dfrac{{C}_{2}{\alpha}_{2}^{2}}{1 - {\alpha}_{2}} . . . \left(4\right)$

eq. \left(1\right) / \left(3\right)$</p><p>$⇒ \dfrac{{\alpha}_{1}}{{\alpha}_{2}} = \dfrac{1}{3}$</p><p>eq \left(2\right) / \left(4\right)$

$⇒ 1 = \dfrac{{C}_{1}{\alpha}_{1}^{2}\left(1 - {\alpha}_{2}\right)}{\left(1 - {\alpha}_{1}\right){C}_{2}{\alpha}_{2}^{2}}$

$\dfrac{{\alpha}_{1}^{2}\left(1 - {\alpha}_{2}\right)}{{\alpha}_{2}^{2}\left(1 - {\alpha}_{1}\right)} = \dfrac{{C}_{2}}{{C}_{1}}$

${C}_{1} = {C}_{2}20y$

$\dfrac{1}{2} = \dfrac{{C}_{2}}{{C}_{1}}$

$\dfrac{{\alpha}_{1}^{2}\left(1 - {\alpha}_{2}\right)}{{\alpha}_{2}^{2}\left(1 - {\alpha}_{1}\right)} = \dfrac{1}{20}$

$\dfrac{20{\alpha}_{1}^{2}}{\left(1 - {\alpha}_{1}\right)} = {\alpha}_{2}^{2}\left(1 - {\alpha}_{2}\right)$

$\dfrac{20}{\left(1 - {\alpha}_{1}\right)} = \dfrac{9}{\left(1 - 3{\alpha}_{1}\right)}$

$20 - 60{\alpha}_{1} = 9 - 9{\alpha}_{1}$

$11 = 51{\alpha}_{1}$

${\alpha}_{1} = \dfrac{11}{51} = 0.22$

$y = 4{\alpha}_{1} = 4 \times \dfrac{11}{51} = 0 .86$