(a) State Raoult's law for a solution containing volatile components. How does Raoult's law become a special case of Henry's law?
(b) $1.00 g$ of a non-electrolyte solute dissolved in $50 g$ of benzene lowered the freezing point of benzene by $0.40 K$ . Find the molar mass of the solute. ( $K_{f}$ for $b e n z e n e = 5.12 K k g / m o l)$

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Solution

(a). According to Raoult's law, vapour pressure of a component is directly proportional to its mole fraction at a particular temperature.
$P_{a}$ = $P_{a}^{o} X_{a}$
So, $P_{b}$ = $P_{b}^{o} X_{b}$
$P_{t o t a l} = P_{a}^{o} X_{a} + P_{b}^{o} X_{b}$
In Henry's law, mole fraction of a gas at a particular temperature is proportional to the pressure exerted over the gas.
$P_{g a s} = k X_{g a s}$
So, we can say that for volatile substance it is a special case of Henry's law.
(b). $Δ T_{f} = K_{f} m$
= $Δ T_{f} = K_{f} \times \frac{W_{b}}{M_{b}} \times \frac{1000}{W_{a}}$
$0.40 = 5.12 \times \frac{1}{M_{b}} \times \frac{1000}{50}$
$M_{b}$ = 256 g/mol

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