Question

State Raoult’s law.
The vapour pressure of pure Benzene at a certain temperature is $0.850bar$. A non-volatile, non-electrolyte solid weighing $0.5g$ when added to $39.0g$ of benzene (molar mass $78gmo{l^{ - 1}}$ ). Vapour pressure of the solution, then is $0.845bar$. What is the molar mass of the solid substance?

Answer 1

Raoult’s law gives the partial pressure of each component in a mixture of components. It gives a relation between partial pressure of each component and the vapor pressure of pure component and mole fraction in a mixture.

Complete step by step answer:
Raoult’s law was given by a French chemist in $1887$ by François-Marie Raoult. It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.
As a consequence of this law it was observed the relative lowering of vapor pressure of a dilute solution containing a nonvolatile solute is equal to the mole fraction of the solute in the solution.
Mathematically Raoult’s law is expressed as
${P_i} = {P_i}^*{\chi _i}$
Where${P_i}$ , ${P_i}^*$ and ${\chi _i}$ are the partial pressure, vapor pressure of the pure component and the mole fraction of the component $i$ respectively.
When a non-volatile solute is added to a pure solvent, this results in a decrease of the vapor pressure of the component. This is a colligative property known as relative lowering of vapor pressure. The relative lowering of vapor pressure is related to the molecular weight of the pure solvent and the weight of the solute and the solvent by the following relation:
$\dfrac{{P - Ps}}{P} = \dfrac{{wM}}{{Wm}}$
Where $P$ is the vapor pressure of the pure solvent, ${P_s}$ is vapor pressure of the solution, $w$ is the mass of the solute, $M$ is the molar mass of the solvent, $W$ is the mass of the solvent and $m$ is the molar mass of the solute.
Given $P = 0.85bar$ , ${P_s} = 0.845bar$ , $w = 0.5g$ , $M = 78g/mol$ , $W = 39.0g$ , $w = ?$
Inserting the values in the equation,
$\dfrac{{0.85 - 0.845}}{{0.85}} = \dfrac{{0.5 \times 78}}{{39 \times w}}$
$w = \dfrac{{0.5 \times 78}}{{0.0058 \times 39}} = 170g/mol.$

Hence, the molar mass of the solid substance is $170g/mol$ .

Note: The added solute takes some of the surface area of the solution. This results in less number of solvent molecules to go to the vapor phase. Hence the relative lowering of vapor pressure occurs by adding non volatile solute to pure solvent.