Q25 / 150 · MHT CET 2026 (15 April Shift 1) · Chemistry
The molar conductivity of $0.01$ M monobasic acid at $25\,^\circ\text{C}$ is $15$ ohm$^{-1}$ cm$^2$ mol$^{-1}$. The molar conductivity of the same acid at infinite dilution is $375$ ohm$^{-1}$ cm$^2$ mol$^{-1}$. Calculate $[\text{H}^+]$ in solution.
- A$2.0 \times 10^{-4}$ M
- B$3.0 \times 10^{-4}$ M
- Correct answerC$4.0 \times 10^{-4}$ M
- D$5.0 \times 10^{-4}$ M
Solution
The degree of dissociation $\alpha$ is given by the ratio of molar conductivity at a given concentration to the molar conductivity at infinite dilution.
$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty}$
Substituting the given values:
$\alpha = \dfrac{15}{375} = 0.04$
For a weak monobasic acid, the concentration of hydrogen ions is given by:
$[\text{H}^+] = C\alpha$
Substituting $C = 0.01$ M and $\alpha = 0.04$:
$[\text{H}^+] = 0.01 \times 0.04 = 4.0 \times 10^{-4}$ M
Answer: $4.0 \times 10^{-4}$ M
$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty}$
Substituting the given values:
$\alpha = \dfrac{15}{375} = 0.04$
For a weak monobasic acid, the concentration of hydrogen ions is given by:
$[\text{H}^+] = C\alpha$
Substituting $C = 0.01$ M and $\alpha = 0.04$:
$[\text{H}^+] = 0.01 \times 0.04 = 4.0 \times 10^{-4}$ M
Answer: $4.0 \times 10^{-4}$ M
Hence, the correct answer is option: 'C'.