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Q.Henry's constant for CH3_3Br(g)_{(g)} is 0.159 mol dm−3^{-3}.bar−1^{-1} at 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0.164 bar.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2024Subjective· 2mImportance★★★★★
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Henry's law: C=KHp=0.159×0.164≈0.0261 mol dm−3C=K_H p = 0.159\times0.164 \approx 0.0261\ mol\,dm^{-3}.

By Henry's law, the solubility (concentration) of a gas in a liquid is directly proportional to its partial pressure above the liquid: C=KH×pC = K_H \times p

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