Henry’s constant for CH3Br(g) is 0·159 mol dm–3.bar–1 at 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0·164 bar.

Q.11 Henry’s constant for CH₃Br(g) is 0·159 mol dm^–3.bar^–1 at 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0·164 bar.

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: S = 0.0261 mol dm\(^{-3}\)

Henry's law: \(S = K_H \times P\)

\(S = 0.159\ \mathrm{mol\,dm^{-3}\,bar^{-1}} \times 0.164\ \mathrm{bar} = 0.02608\ \mathrm{mol\,dm^{-3}}\)

Solubility of CH\(_3\)Br in water at 25 °C \(\approx 2.61 \times 10^{-2}\) mol dm\(^{-3}\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Henry’s constant for CH3Br(g) is 0·159 mol dm–3.bar–1 at 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0·164 bar.

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