Question

Calculate the mass of a non-volatile solute (molar mass 40 ${\text{g/mol}}$) which can be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Answer 1

This question can be solved from the knowledge of relative lowering of vapour pressure of a solvent which is a colligative property of solutions. We shall substitute the values in the formula given to find the mass required.
Formula used: $\dfrac{{{{\text{p}}^{\text{0}}} - {\text{p}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_{\text{1}}}}}$
Where ${{\text{p}}^0}$ the initial pressure, p is the final pressure, ${{\text{w}}_{\text{2}}}$ is the weight of the solute, ${{\text{w}}_{\text{1}}}$ is the weight of the solvent, ${{\text{M}}_{\text{1}}}$ and ${{\text{M}}_{\text{2}}}$ are the molecular weights of the solvent, and the solute respectively.

Complete step by step answer:
According to the formula, relative lowering of vapour pressure, $\dfrac{{{{\text{p}}^{\text{0}}} - {\text{p}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_{\text{1}}}}}$
Let the initial vapour pressure of octane be p.
Then after the addition of the solute, the vapour pressure becomes, $\dfrac{{80}}{{100}}{\text{p}}$ = $0.8{\text{p}}$
${{\text{M}}_{\text{1}}}$ and ${{\text{M}}_{\text{2}}}$ = 114g and 40 g, respectively.
The mass of octane taken,${{\text{w}}_{\text{1}}}$ = 114 g
Putting all the values, we get,
$\dfrac{{{\text{p}} - {\text{0}}{\text{.8p}}}}{{\text{p}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {\text{114}}}}{{{\text{40}} \times {\text{114}}}}$
Solving, for ${{\text{w}}_{\text{2}}}$ we get:
${{\text{w}}_2} = 8{\text{ g}}$

So the mass of solute in octane is 8 g.

Additional information:
There are three other types of colligative properties of a solution: lowering of boiling point, depression of freezing point, and osmotic pressure of the solution. The relative lowering of vapour pressure of a solution is based on the Raoult’s Law which states that the relative lowering of the vapour pressure of the solution is equal to the mole fraction of the solute in the solution.

Note:
The pressure exerted on the walls of the container by the vapour of a solvent is called its vapour pressure and this property of the solvent is lowered by the presence of solute molecules when a solution is formed from the solvent. This property is a colligative property which depends only on the number of the solute particles, irrespective of the nature of the particles.