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.
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Solution and Explanation
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: \[ C = 0.159 \times 0.164 = 0.0261 \, \text{mol dm}^{-3} \]
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