Q119 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Chemistry
The degree of dissociation of a monobasic acid in its 0.1M solution at 300 K is $1.342 \times 10^{-2}$. Calculate the degree of dissociation of the same acid in its 0.02M solution.
- A$1.0 \times 10^{-2}$
- B$2.0 \times 10^{-2}$
- Correct answerC$3.0 \times 10^{-2}$
- D$4.0 \times 10^{-2}$
Solution
Using Ostwald's dilution law for a weak monobasic acid:
$K_a = C \alpha^2$
For the first solution, $C_1 = 0.1$ M and $\alpha_1 = 1.342 \times 10^{-2}$:
$K_a = 0.1 \times (1.342 \times 10^{-2})^2 = 0.1 \times 1.8 \times 10^{-4} = 1.8 \times 10^{-5}$
For the second solution, $C_2 = 0.02$ M:
$K_a = C_2 \alpha_2^2$
$1.8 \times 10^{-5} = 0.02 \times \alpha_2^2$
$\alpha_2^2 = \dfrac{1.8 \times 10^{-5}}{0.02} = 9 \times 10^{-4}$
$\alpha_2 = 3.0 \times 10^{-2}$
Answer: $3.0 \times 10^{-2}$
$K_a = C \alpha^2$
For the first solution, $C_1 = 0.1$ M and $\alpha_1 = 1.342 \times 10^{-2}$:
$K_a = 0.1 \times (1.342 \times 10^{-2})^2 = 0.1 \times 1.8 \times 10^{-4} = 1.8 \times 10^{-5}$
For the second solution, $C_2 = 0.02$ M:
$K_a = C_2 \alpha_2^2$
$1.8 \times 10^{-5} = 0.02 \times \alpha_2^2$
$\alpha_2^2 = \dfrac{1.8 \times 10^{-5}}{0.02} = 9 \times 10^{-4}$
$\alpha_2 = 3.0 \times 10^{-2}$
Answer: $3.0 \times 10^{-2}$
Hence, the correct answer is option: 'C'.