Question

What is vapour pressure of a solution containing 1 mol of a non-volatile solute in 36 g of water? ($P_1^0 = 400\ \text{mm Hg}$)

• A. 334 mm Hg
• B. 267 mm Hg
• C. 240 mm Hg
• D. 284 mm Hg

Hint:

An alternative time-saving path is using Raoult's Law rewritten directly for the solvent's perspective: $P_{\text{soln}} = x_{\text{solvent}} \times P^0$. Here, $x_{\text{solvent}} = \frac{2}{3}$, so $P_{\text{soln}} = \frac{2}{3} \times 400 = \frac{800}{3} = 267\ \text{mm Hg}$ in just one single step!

Answer 1

The Correct Option is B

Step 1: Identify the colligative law in play.
With a non-volatile solute, Raoult's law says the solution pressure is the pure solvent pressure scaled by the solvent's mole fraction: $P_1 = x_1 P_1^0$.
Step 2: List the data.
Solute $= 1$ mol, water $= 36$ g, and $P_1^0 = 400$ mm Hg.
Step 3: Convert water mass to moles.
Water's molar mass is 18, so moles of water $= \frac{36}{18} = 2$ mol.
Step 4: Find the solvent mole fraction.
$x_1 = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{solute}}} = \frac{2}{2 + 1} = \frac{2}{3}$.
Step 5: Apply the law directly.
$P_1 = x_1 P_1^0 = \frac{2}{3} \times 400$.
Step 6: Do the arithmetic.
$P_1 = \frac{800}{3} \approx 266.7$ mm Hg.
Step 7: Round to the option.
This rounds to 267 mm Hg, which is option (2).
\[ \boxed{P_1 = \tfrac{2}{3} \times 400 \approx 267\ \text{mm Hg}} \]