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Henry's Law

a On dissolvi...

Question

a) On dissolving 2.34 g of solute in 40 g of benzene the boiling point of solution was higher than that of benzene by $0.81 K, K_{b}$ value for benzene is 2.53 K kg $m o l^{- 1}$ . Calculate the molar mass of the solute.
b) State Henry's law. Write its mathematical form.

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Solution

a)
$M_{2} = \frac{K_{b} \times W_{2}}{Δ T_{b} W_{1}}$
$M_{2} = \frac{2.53 \times 2.34}{0.81 \times 40}$
$M_{2} = 0.183$ kg/mol
$M_{2} = 183$ g/mol
b)
Henry's law: Solubility of a gas in a liquid at constant temperature is proportional to the pressure of the gas above the solution.
$S = K P$
$S =$ solubility of the gas in mol/L
$K =$ Henry's law constant with units mol/ L / atm.
$P =$ pressure of the gas in atmosphere.

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