The Henry's Law constant of methyl bromide,CH₃Br,is k = 0.159 mol/(L ∙ atm)at 25°C.What is the solubility of methyl bromide in water at 25°C and at a partial pressure of 250.mm Hg? A)0.0523 mol/L B)0.329 mol/L C)0.483 mol/L D)39.8 mol/L

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Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid. The formula for Henry's Law is S = kP, where S is the solubility of the gas (in mol/L), k is the Henry's Law constant (in mol/(L·atm)), and P is the partial pressure of the gas (in atm). Given that the Henry's Law constant, k, for methyl bromide is 0.159 mol/(L·atm) and the partial pressure is 250 mm Hg, which is equivalent to 250 mm Hg * (1 atm / 760 mm Hg) = 0.3289 atm, the solubility, S, can be calculated as follows: S = 0.159 mol/(L·atm) * 0.3289 atm = 0.0523 mol/L.

The Henry's Law Constant of Methyl Bromide,CH3Br,is K = 0

The Henry's Law Constant of Methyl Bromide,CH₃Br,is K = 0