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Question
Henry’s constant for CH3Br(g) is 0.159 mol dm−3 bar−1 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0.164 bar.
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Solution
Given: Henry’s constant for CH3Br(g) = KH = 0.159 mol dm−3 bar−1
Partial pressure (P) = 0. 164 bar
To find: Solubility (S) = ?
Formula: S = KHP
S = 0.159 × 0.164
S = 0.02607 mol L−1
∴ Solubility of CH3Br(g) at the given condition is 0.02607 ≅ 0.0261 mol L−1.
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