Question

Calculate the mass of a non-volatile solute (molar mass $40\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$) which should be dissolved in ${\text{114}}\,{\text{g}}$octane to reduce its vapour pressure to $80\,\%$

Answer 1

A non-volatile substance is a substance that does not evaporate into a gas under given conditions. Non-volatile substances have a low vapor pressure and a high boiling point. A volatile substance sublimates at a room temperature and they have higher vapor pressures.

Complete step by step answer:
Given that,
Molar mass, ${{\text{M}}_{\text{2}}}$ =$40\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Amount of octane, ${{\text{w}}_{\text{1}}}$ = ${\text{114}}\,{\text{g}}$
Vapour pressure reduced to = $80\,\%$
Let us take the vapour pressure of octane to be ${{\text{p}}_{\text{1}}}^{\text{0}}$
Now, vapour pressure octane after it is dissolved in the non-volatile solute is $\dfrac{{80}}{{100}}{{\text{p}}_{\text{1}}}^{\text{0}} = \,0.8\,{{\text{p}}_{\text{1}}}^{\text{0}}$
Now we will calculate the molar mass of octane.
Molar mass of octane $\left( {{{\text{C}}_{\text{8}}}{{\text{H}}_{{\text{18}}}}} \right),\,{{\text{M}}_{\text{1}}} = \,8 \times 12 + 8 \times 1 = \,114\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Now, we apply the relation from Raoult’s law,

$$\dfrac{{\left( {{p_1}^0 - {p_1}} \right)}}{{{p_1}^0}} = \dfrac{{\left( {{w_2} \times {M_1}} \right)}}{{\left( {{M_2} \times {w_1}} \right)}} \\\ \Rightarrow \,\dfrac{{\left( {{p_1}^0 - 0.8\,{p_1}^0} \right)}}{{{p_1}^0}} = \,\dfrac{{\left( {{w_2} \times 114} \right)}}{{\left( {40 \times 114} \right)}} \\\ \Rightarrow \,\dfrac{{0.2\,{p_1}^0}}{{{p_1}^0}} = \,\dfrac{{{w_2}}}{{40}} \\\ \Rightarrow {w_2} = \,8\,g \\\$$

The required mass of the solute is ${\text{8}}\,{\text{g}}$.

Additional Information:
Raoult’s law states that partial vapour pressure of a solvent in a solution or in mixture is equal or to the vapour pressure of the pure solvent which is multiplied by its mole fraction in the solution.
Raoult’s law equation can be written as;
${{\text{P}}_{{\text{solution}}}} = \,{{\text{X}}_{{\text{Solvent}}}}{{\text{P}}^{\text{0}}}_{{\text{solvent}}}$
Here,
${{\text{P}}_{{\text{solution}}}}$ = vapour pressure of the solution
${{\text{X}}_{{\text{Solvent}}}}$= mole fraction of the solvent
${{\text{P}}^{\text{0}}}_{{\text{solvent}}}$= vapour pressure of the pure solvent
Few limitations to Raoult’s law are:
1)Raoult's law is useful for describing ideal solutions. However, ideal solutions are rare and hard to find.
2)The negative deviation is seen when the vapour pressure is lower than that of the expected from Raoult's law. A positive deviation is seen when the cohesion between the similar molecules is greater or exceeds the adhesion between dissimilar molecules. Both components of the mixture can easily escape from the solution.

Note:
Raoult's law is valid only in ideal solutions. In an ideal solution, the solvent-solute interaction is the same as that of the solvent - solvent or solute - solute interaction. This tells that both the solute and the solvent take the same amount of energy in order to escape to the vapour phase when they exist in their pure states.