Question

The mass of a non-volatile solute of molar mass$40\mathrm{g}/\mathrm{mol}$ that should be dissolved in $114\mathrm{g}$ of octane to lower its vapour pressure by $20\%$ is:

Answer 1

Step 1: The mole fraction of non-volatile solute decreases proportionally to the relative decrease in vapour pressure.

• Let w g of the non-volatile solute is used. Its molar mass is $40\mathrm{g}{\mathrm{mol}}^{-1}$
• The number of moles of non-volatile solute $=\frac{\mathrm{w}}{40}=0.025\mathrm{w}$
• $114\mathrm{g}$ of octane (molar mass $114\mathrm{g}{\mathrm{mol}}^{-1}$) corresponds to$\frac{114}{114}=1\mathrm{mole}$

Step 2: Applying formulae for mole fraction of non-volatile solute

• The mole fraction of non-volatile solute is =$\frac{1+0.025\mathrm{w}}{0.025\mathrm{w}}$
• The mole fraction of the nonvolatile solute decreases proportionally to the relative decrease in vapor pressure.
• $\frac{\mathrm{P}°-\mathrm{P}}{\mathrm{P}°}={\mathrm{X}}_{\mathrm{B}}$
• $$\begin{gathered} 100(0.025\mathrm{w})=20(1+0.025\mathrm{w}) \\ 2.5\mathrm{w}=20+0.50\mathrm{w} \\ 2\mathrm{w}=20 \\ \mathrm{w}=10\mathrm{g} \end{gathered}$$

The mass of a non-volatile solute of molar mass $40\mathrm{g}{\mathrm{mol}}^{-1}$ that should be dissolved in $114\mathrm{g}$ of octane to lower its vapor pressure by $20\%$ is $10\mathrm{g}$.