Molar Conductivity Value of Monobasic Acid Solution

To find the value of $y$, we utilize Ostwald's dilution law for the weak monobasic acid and the relation between molar conductivity ($\Lambda_m$), limiting molar conductivity ($\Lambda_m^\circ$), and degree of dissociation ($\alpha$).

Step 1: Express $\alpha$ and $\alpha'$ in terms of $y$

The degree of dissociation $\alpha$ is given by: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

For the initial solution: $\Lambda_m = y \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}$ $\Lambda_m^\circ = 4 \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}$ $\alpha = \frac{y \times 10^2}{4 \times 10^2} = \frac{y}{4} \implies y = 4\alpha$

When the solution is diluted $20$ times, the concentration becomes $C' = \frac{C}{20}$, and the molar conductivity increases to: $\Lambda_m' = 3y \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}$

The new degree of dissociation $\alpha'$ is: $\alpha' = \frac{\Lambda_m'}{\Lambda_m^\circ} = \frac{3y \times 10^2}{4 \times 10^2} = \frac{3y}{4} = 3\alpha$

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Step 2: Apply Ostwald's Dilution Law

The acid dissociation constant ($K_a$) remains unchanged at constant temperature ($298\text{ K}$):

1. Before dilution: $K_a = \frac{C \alpha^2}{1 - \alpha}$

2. After dilution: $K_a = \frac{C' (\alpha')^2}{1 - \alpha'} = \frac{\left(\frac{C}{20}\right) (3\alpha)^2}{1 - 3\alpha} = \frac{9 C \alpha^2}{20 (1 - 3\alpha)}$

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Step 3: Solve for $\alpha$ and $y$

Equating the two expressions for $K_a$: $\frac{C \alpha^2}{1 - \alpha} = \frac{9 C \alpha^2}{20 (1 - 3\alpha)}$

Since $\alpha \neq 0$, we can simplify to: $1 - \alpha = \frac{20 (1 - 3\alpha)}{9}$

$9(1 - \alpha) = 20(1 - 3\alpha)$

$9 - 9\alpha = 20 - 60\alpha$

$51\alpha = 11$

$\alpha = \frac{11}{51}$

Now, substitute $\alpha$ to find $y$: $y = 4\alpha = 4 \times \frac{11}{51} = \frac{44}{51} \approx 0.86$

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Final Answer:

The value of $y$ is $0.86$ (or $\frac{44}{51}$).