Question

What is degree of dissociation of $CH_3COOH$ if, $\Lambda^\circ (CH_3COO^-) = 50\text{ S cm}^2\text{ mol}^{-1}$, $\Lambda^\circ (H^+) = 350\text{ S cm}^2\text{ mol}^{-1}$ and molar conductivity of $5 \times 10^{-2}\text{ M }CH_3COOH$ is $20\text{ S cm}^2\text{ mol}^{-1}$ ?

• A. $1.25 \times 10^{-4}$
• B. $5 \times 10^{-2}$
• C. $1.25 \times 10^{-2}$
• D. $5 \times 10^{-4}$

Hint:

$\alpha$ is the ratio of conductivity at a given concentration to that at infinite dilution.

Answer 1

The Correct Option is B

Step 1: Kohlrausch's Law
$\Lambda^\circ (CH_3COOH) = \Lambda^\circ (CH_3COO^-) + \Lambda^\circ (H^+) = 50 + 350 = 400\text{ S cm}^2\text{ mol}^{-1}$.

Step 2: Formula
Degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$.

Step 3: Calculation
$\alpha = \frac{20}{400} = \frac{2}{40} = \frac{1}{20}$.
$\alpha = 0.05 = 5 \times 10^{-2}$.

Step 4: Conclusion
The degree of dissociation is $5 \times 10^{-2}$.
Final Answer: (B)