Question

Find the mass of non-volatile, non-electrolyte solute (molar mass 50 g/mol) needed to be dissolved in 114 g of octane to reduce its vapour pressure to 75%:
(A) 150 g
(B) 75 g
(C) 37.5 g
(D) 50 g

Answer 1

Relative lowering of vapour pressure is a colligative property as it is independent of the nature of the solute but is dependent on the moles of that solute.
The relation between the relative lowering of vapour pressure of a solution and the mole fraction of the non-volatile solute in the solution is given by Raoult’s law:
$\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{2}}}}}{{{{\text{n}}_{\text{1}}}{\text{ + }}{{\text{n}}_{\text{2}}}}}$
where ${{\text{p}}^{\text{0}}}$ denotes the vapour pressure of the pure solvent, ${{\text{p}}_{\text{s}}}$ denotes the vapour pressure of the solution, ${{\text{n}}_{\text{2}}}$ denotes the number of moles of the solute and ${{\text{n}}_1}$ denotes the number of moles of the solvent.

Complete step by step solution:
Given, the molar mass of a non-volatile, non-electrolyte solute $= 50$ gram per mole.
Also given, the mass of octane ${\text{ = 114g}}$ .
Also, vapour pressure is reduced to 75 percent.
The term ${{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}$ represent the lowering of vapour pressure and $\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}$ represent the relative lowering of vapour pressure and $\dfrac{{{{\text{n}}_{\text{2}}}}}{{{{\text{n}}_{\text{1}}}{\text{ + }}{{\text{n}}_{\text{2}}}}}$ represent the mole fraction of that solute.
Therefore, we can write:
$\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}}{\text{/}}{{\text{M}}_{\text{2}}}}}{{{{\text{w}}_{\text{2}}}{\text{/}}{{\text{M}}_{\text{2}}}{\text{ + }}{{\text{w}}_{\text{1}}}{\text{/}}{{\text{M}}_{\text{1}}}}}$
where ${{\text{w}}_{\text{2}}}$ and ${{\text{M}}_{\text{2}}}$ represent the mass in grams and molar mass of the solute respectively and ${{\text{w}}_1}$ and ${{\text{M}}_1}$ represent the mass in grams and molar mass of the solvent respectively.
Here, we need to find out the mass of the non-volatile, non-electrolyte solute $\left( {{{\text{w}}_{\text{2}}}} \right)$ .
Let the vapour pressure of the solvent $\left( {{{\text{p}}^{\text{0}}}} \right)$ be 100.
Since the vapour pressure will be 75 percent of 100, i.e., 75, so,
${{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}{\text{ = 75}}$
By question,
${{\text{M}}_{\text{2}}}{\text{ = 50g/mol,}}{{\text{w}}_{\text{1}}}{\text{ = 114g}}$ $$
Molar mass of octane $\left( {{{\text{C}}_{\text{8}}}{{\text{H}}_{{\text{18}}}}} \right)$ ,
${{\text{M}}_{\text{1}}}{{ = 12 \times 8 + 1 \times 18}} \\\ \Rightarrow {{\text{M}}_{\text{1}}}{\text{ = 114g}} \\\$
Using the formula for relative lowering of vapour pressure, we can write:
$\dfrac{{{\text{75}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}}{\text{/50}}}}{{{{\text{w}}_{\text{2}}}{\text{/50 + 114/114}}}} \\\ \Rightarrow {\text{0}}{\text{.75 = }}\dfrac{{{{\text{w}}_{\text{2}}}{\text{/50}}}}{{{{\text{w}}_{\text{2}}}{\text{/50 + 1}}}} \\\ \Rightarrow {{\text{w}}_{\text{2}}}{\text{/50 + 1 = }}{{\text{w}}_{\text{2}}}{{/50 \times 0}}{\text{.75}} \\\ \Rightarrow {{\text{w}}_{\text{2}}}{\text{ + 50 = }}\dfrac{{{{\text{w}}_{\text{2}}}}}{{{\text{0}}{\text{.75}}}} \\\ \Rightarrow {\text{0}}{\text{.75}}{{\text{w}}_{\text{2}}}{\text{ + 37}}{\text{.5 = }}{{\text{w}}_{\text{2}}} \\\ \Rightarrow {\text{0}}{\text{.25}}{{\text{w}}_{\text{2}}}{\text{ = 37}}{\text{.5}} \\\ \Rightarrow {{\text{w}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{37}}{\text{.5}}}}{{{\text{0}}{\text{.25}}}}{\text{ = 150g}} \\\$
Therefore, the mass of the non-volatile, non-electrolyte solute is 150 g.
So, the correct option is option (A).

Note:
An alternative formula has been derived to be used for all solutions, whether they are dilute or concentrated. This is given by:
$\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}_{\text{s}}}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{2}}}}}{{{{\text{n}}_{\text{1}}}}} \\\ \Rightarrow \dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}_{\text{s}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}}{{\text{M}}_{\text{1}}}}}{{{{\text{w}}_{\text{1}}}{{\text{M}}_{\text{2}}}}} \\\$
where ${{\text{w}}_{\text{2}}}$ and ${{\text{M}}_{\text{2}}}$ represent the mass in grams and molar mass of the solute respectively and ${{\text{w}}_1}$ and ${{\text{M}}_1}$ represent the mass in grams and molar mass of the solvent respectively.