The value of y is ___.

The limiting molar conductivity (at infinite dilution) of the weak monobasic acid is given as $$\Lambda_\infty = 4 \times 10^2\; S\,cm^{2}\,mol^{-1}=400\;S\,cm^{2}\,mol^{-1}$$.

For the initial solution at concentration $$c$$:
  • molar conductivity $$\Lambda_1 = y \times 10^2 = 100y\;S\,cm^{2}\,mol^{-1}$$.
  • degree of dissociation $$\alpha_1 = \dfrac{\Lambda_1}{\Lambda_\infty} = \dfrac{100y}{400} = \dfrac{y}{4}$$.

After 20-fold dilution the concentration becomes $$\dfrac{c}{20}$$ and
  • molar conductivity $$\Lambda_2 = 3y \times 10^2 = 300y\;S\,cm^{2}\,mol^{-1}$$.
  • degree of dissociation $$\alpha_2 = \dfrac{\Lambda_2}{\Lambda_\infty} = \dfrac{300y}{400} = \dfrac{3y}{4}$$.

For a weak acid $$HA \rightleftharpoons H^+ + A^-$$, the dissociation constant is

$$K_a = \dfrac{c\,\alpha^2}{1-\alpha}\;-(1)$$

Using $$(1)$$ for the two solutions (same acid, so same $$K_a$$):
  Initial: $$K_a = \dfrac{c\left(\dfrac{y}{4}\right)^2}{1-\dfrac{y}{4}}$$.
  Diluted: $$K_a = \dfrac{\dfrac{c}{20}\left(\dfrac{3y}{4}\right)^2}{1-\dfrac{3y}{4}}$$.

Equate the two expressions and cancel common factors:

$$\dfrac{c\,\dfrac{y^{2}}{16}}{1-\dfrac{y}{4}} \;=\; \dfrac{\dfrac{c}{20}\,\dfrac{9y^{2}}{16}}{1-\dfrac{3y}{4}}$$

$$\Rightarrow \dfrac{1}{1-\dfrac{y}{4}} \;=\; \dfrac{9}{20}\,\dfrac{1}{1-\dfrac{3y}{4}}$$

Cross-multiplying:

$$1-\dfrac{3y}{4} \;=\; \dfrac{9}{20}\Bigl(1-\dfrac{y}{4}\Bigr)$$

Multiply by 20 to clear the denominator:

$$20\Bigl(1-\dfrac{3y}{4}\Bigr) = 9\Bigl(1-\dfrac{y}{4}\Bigr)$$

$$20 - 15y = 9 - \dfrac{9y}{4}$$

Bring all terms to one side:

$$20 - 15y - 9 + \dfrac{9y}{4} = 0$$

$$11 - \dfrac{51y}{4} = 0$$

$$\dfrac{51y}{4} = 11 \quad\Rightarrow\quad y = \dfrac{44}{51} \approx 0.86$$

Thus

The required value of $$y$$ is approximately $$0.86$$, which lies in the range 0.80 - 0.90.