Question

The solubility of \(N_2(g)\) in water exposed to the atmosphere, when the partial pressure is \(593\,\text{mm}\), is \(5.3 \times 10^{-4}\,\text{M}\). Its solubility at \(760\,\text{mm}\) and at the same temperature is

• A. \(4.1 \times 10^{-4}\,\text{M}\)
• B. \(6.8 \times 10^{-4}\,\text{M}\)
• C. \(1500\,\text{M}\)
• D. \(2400\,\text{M}\)

Hint:

Henry’s law: \[ S \propto P \] If pressure increases, solubility of a gas in a liquid also increases proportionally (at constant temperature).

Answer 1

The Correct Option is B

Step 1: According to Henry’s law, at constant temperature, the solubility of a gas in a liquid is directly proportional to its partial pressure. \[ S \propto P \]
Step 2: Hence, \[ \frac{S_1}{P_1} = \frac{S_2}{P_2} \] Given: \[ S_1 = 5.3 \times 10^{-4}\,\text{M}, \quad P_1 = 593\,\text{mm}, \quad P_2 = 760\,\text{mm} \]
Step 3: Calculate the new solubility: \[ S_2 = S_1 \times \frac{P_2}{P_1} = 5.3 \times 10^{-4} \times \frac{760}{593} \] \[ S_2 \approx 5.3 \times 10^{-4} \times 1.28 = 6.8 \times 10^{-4}\,\text{M} \]
Step 4: Therefore, the solubility of nitrogen at \(760\,\text{mm}\) pressure is: \[ \boxed{6.8 \times 10^{-4}\,\text{M}} \]