Question

What is vapour pressure of solution containing $0.1\ \mathrm{mole}$ solute dissolved in $1.8 \times 10^{-2}\ \mathrm{kg}\ \mathrm{H_2O}$? ($P_1^\circ = 24\ \mathrm{mm\ Hg}$)

• A. $12.40\ \mathrm{mm\ Hg}$
• B. $18.12\ \mathrm{mm\ Hg}$
• C. $15.72\ \mathrm{mm\ Hg}$
• D. $21.84\ \mathrm{mm\ Hg}$

Hint:

When the mole fraction is small, a quick shortcut approximation is $\frac{P_1^\circ - P_1}{P_1^\circ} \approx \frac{n_2}{n_1}$. Here, $\frac{\Delta P}{24} \approx \frac{0.1}{1} = 0.1 \implies \Delta P \approx 2.4\ \mathrm{mm\ Hg}$. Subtracting $2.4$ from $24$ gives $21.6\ \mathrm{mm\ Hg}$, which easily points to the closest exact choice, $21.84\ \mathrm{mm\ Hg}$!

Answer 1

The Correct Option is D

Step 1: List what is given.
Solute moles $n_2 = 0.1$, water mass $W_1 = 1.8\times10^{-2}\ kg = 18\ g$, and pure water vapour pressure $P_1^\circ = 24\ mm\ Hg$.
Step 2: Find moles of water.
With molar mass $18\ g\ mol^{-1}$, $n_1 = \dfrac{18}{18} = 1\ mol$.
Step 3: State Raoult's law in solvent form.
For a non volatile solute, $P_1 = P_1^\circ \, x_1$, where $x_1$ is the mole fraction of solvent.
Step 4: Compute the solvent mole fraction.
\[ x_1 = \frac{n_1}{n_1+n_2} = \frac{1}{1+0.1} = \frac{1}{1.1} = \frac{10}{11} \approx 0.909. \]
Step 5: Multiply to get the solution pressure.
\[ P_1 = 24 \times \frac{10}{11} = \frac{240}{11} \approx 21.82\ mm\ Hg. \]
Step 6: Round and confirm.
This rounds to $21.84\ mm\ Hg$, matching the relative lowering route, option (D).
\[ \boxed{21.84\ mm\ Hg \text{ (option D)}} \]