Q15 / 150 · MHT CET 2026 (11 April Shift 1) · Chemistry
The solubility of $\text{N}_2$ gas in water at $25^\circ$C and $1$ bar is $6.85 \times 10^{-4} \text{ mol L}^{-1}$. Calculate the solubility of $\text{N}_2$ gas in water at the same temperature when the partial pressure of $\text{N}_2$ is $0.70$ bar
- A$5.138 \times 10^{-4} \text{ mol L}^{-1}$
- B$5.480 \times 10^{-4} \text{ mol L}^{-1}$
- Correct answerC$4.795 \times 10^{-4} \text{ mol L}^{-1}$
- D$4.875 \times 10^{-4} \text{ mol L}^{-1}$
Solution
According to Henry's law, the solubility of a gas in a liquid is directly proportional to its partial pressure at a constant temperature.
$S \propto P$
$\dfrac{S_1}{P_1} = \dfrac{S_2}{P_2}$
Given $S_1 = 6.85 \times 10^{-4} \text{ mol L}^{-1}$, $P_1 = 1 \text{ bar}$, and $P_2 = 0.70 \text{ bar}$.
$S_2 = S_1 \times \dfrac{P_2}{P_1}$
$S_2 = 6.85 \times 10^{-4} \times \dfrac{0.70}{1}$
$S_2 = 4.795 \times 10^{-4} \text{ mol L}^{-1}$
Answer: $4.795 \times 10^{-4} \text{ mol L}^{-1}$
$S \propto P$
$\dfrac{S_1}{P_1} = \dfrac{S_2}{P_2}$
Given $S_1 = 6.85 \times 10^{-4} \text{ mol L}^{-1}$, $P_1 = 1 \text{ bar}$, and $P_2 = 0.70 \text{ bar}$.
$S_2 = S_1 \times \dfrac{P_2}{P_1}$
$S_2 = 6.85 \times 10^{-4} \times \dfrac{0.70}{1}$
$S_2 = 4.795 \times 10^{-4} \text{ mol L}^{-1}$
Answer: $4.795 \times 10^{-4} \text{ mol L}^{-1}$
Hence, the correct answer is option: 'C'.