Q14 / 150 · MHT CET 2026 (11 April Shift 2) · Chemistry
The partial pressure of a gas at $25^\circ$C is $0.18$ atm, calculate the concentration of the gas dissolved at the same temperature,
If $K_H$ is $0.15\ \text{mol dm}^{-3}\ \text{atm}^{-1}$
If $K_H$ is $0.15\ \text{mol dm}^{-3}\ \text{atm}^{-1}$
- Correct answerA$0.027$ M
- B$1.8$ M
- C$4.5$ M
- D$0.45$ M
Solution
According to Henry's law, the concentration of a dissolved gas is directly proportional to its partial pressure.
$C = K_H \times P$
Given:
$P = 0.18 \text{ atm}$
$K_H = 0.15 \text{ mol dm}^{-3} \text{ atm}^{-1}$
Substituting the values:
$C = 0.15 \times 0.18$
$C = 0.027 \text{ mol dm}^{-3}$
Since $1 \text{ mol dm}^{-3} = 1 \text{ M}$, the concentration is $0.027 \text{ M}$.
Answer: $0.027$ M
$C = K_H \times P$
Given:
$P = 0.18 \text{ atm}$
$K_H = 0.15 \text{ mol dm}^{-3} \text{ atm}^{-1}$
Substituting the values:
$C = 0.15 \times 0.18$
$C = 0.027 \text{ mol dm}^{-3}$
Since $1 \text{ mol dm}^{-3} = 1 \text{ M}$, the concentration is $0.027 \text{ M}$.
Answer: $0.027$ M
Hence, the correct answer is option: 'A'.