At $298 \mathrm{~K}$, the limiting molar conductivity of a weak monobasic acid is $4 \times 10^{2}…

$\mathrm{\alpha }=\frac{{\mathrm{\lambda }}_{\mathrm{m}}}{{\mathrm{\lambda }}_{\mathrm{m}}^{\circ }}$

${\mathrm{\alpha }}_1=\frac{\mathrm{y}\times {10}^{+2}}{4\times {10}^2}=\frac{\mathrm{y}}{4} ...\left(1\right)$

${\mathrm{k}}_{\mathrm{a}}=\frac{{\mathrm{C}}_1{\mathrm{\alpha }}_1^2}{1-{\mathrm{\alpha }}_1} ...\left(2\right)$

After dilution

${\mathrm{\alpha }}_2=\frac{3\mathrm{y}\times {10}^2}{4\times {10}^2}=\frac{3\mathrm{y}}{4} ...\left(3\right)$

${\mathrm{k}}_{\mathrm{a}}=\frac{{\mathrm{C}}_2{\mathrm{\alpha }}_2^2}{1-{\mathrm{\alpha }}_2} ...\left(4\right)$

eq. $\left(1\right)/\left(3\right)$

$\Rightarrow \frac{{\mathrm{\alpha }}_1}{{\mathrm{\alpha }}_2}=\frac{1}{3}$

eq $\left(2\right)/\left(4\right)$

$\Rightarrow 1=\frac{{\mathrm{C}}_1{\mathrm{\alpha }}_1^2\left(1-{\mathrm{\alpha }}_2\right)}{\left(1-{\mathrm{\alpha }}_1\right){\mathrm{C}}_2{\mathrm{\alpha }}_2^2}$

$\frac{{\mathrm{\alpha }}_1^2\left(1-{\mathrm{\alpha }}_2\right)}{{\mathrm{\alpha }}_2^2\left(1-{\mathrm{\alpha }}_1\right)}=\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}$

${\mathrm{C}}_1={\mathrm{C}}_2 20\mathrm{y}$

$\frac{1}{2}=\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}$

$\frac{{\mathrm{\alpha }}_1^2\left(1-{\mathrm{\alpha }}_2\right)}{{\mathrm{\alpha }}_2^2\left(1-{\mathrm{\alpha }}_1\right)}=\frac{1}{20}$

$\frac{20{\mathrm{\alpha }}_1^2}{\left(1-{\mathrm{\alpha }}_1\right)}=\frac{{\mathrm{\alpha }}_2^2}{\left(1-{\mathrm{\alpha }}_2\right)}$

$\frac{20}{\left(1-{\mathrm{\alpha }}_1\right)}=\frac{9}{\left(1-{\mathrm{\alpha }}_2\right)}$

$20-60{\mathrm{\alpha }}_1=9-9{\mathrm{\alpha }}_1$

$11=51{\mathrm{\alpha }}_1$

${\mathrm{\alpha }}_1=\frac{11}{51}=0.22$