1. State Raoult’s law for a solution containing volatile componen… - Vidyadip

Question

State Raoult’s law for a solution containing volatile components.

How does Raoult’s law become a special case of Henry’s law?

$1.00\ g$ 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 benzene$= 5.12\ kg\ mol^{-1})$.

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Answer

Partial vapour pressure of a liquid component is directly proportional to its mole fraction in its solution.

The partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant $K_H$ differs from $P^o_A$. Thus, Raoult’s law becomes a special case of Henry’s law in which $K_H$ becomes equal to $P^o_A$.

Given $W_B = 1∙00g$;

$W_A = 50g;$

$K_f = 5∙12K kg/ mol- 1;$

$\triangle T_f = 0∙40K$

$\Delta\text{T}_{f}=\text{K}_{f}\frac{\text{W}_{B}\times{1000}}{\text{M}_{B}\times\text{W}_{A}\text{(in grams)}}$
$\text{M}_{B}=\text{K}_{f}\frac{\text{W}_{B}\times{1000}}{\Delta\text{T}_{f}\times\text{W}_{A}}$
$\text{M}_{B}=\frac{\text{5.12}\times{1}\times{1000}}{\text{0.40}\times\text{50}}$
$= 256g mol^{-1}$

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