Question

Henry’s constant for CH\(_3\)Br(g) is 0.159 mol dm\(^{-3}\) bar\(^{-1}\) at 25\(^{\circ}\)C. Calculate its solubility in water at 25\(^{\circ}\)C if its partial pressure is 0.164 bar.

Hint:

According to Henry's law, the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid.

Answer 1

Henry's law is expressed as follows: \[ C = k_H P \] Definitions: 
- \( C \) represents the solubility of the gas in mol dm\(^{-3}\), 
- \( k_H \) is Henry's constant, here \( 0.159 \, \text{mol dm}^{-3} \text{ bar}^{-1} \), 
- \( P \) stands for the partial pressure of the gas, \( 0.164 \, \text{bar} \). 
Calculating the solubility: \[ C = 0.159 \times 0.164 = 0.0261 \, \text{mol dm}^{-3} \]