Question

Calculate the relative lowering of vapour pressure of solution containing 3 g urea in 50 g water. [molar mass of urea = 60 g mol$^{-1}$ ]}

• A. 0.013
• B. 0.025
• C. 0.018
• D. 0.028

Hint:

For dilute solutions, $\frac{P^\circ - P}{P^\circ} \approx \frac{n_2}{n_1}$.

Answer 1

The Correct Option is C

Step 1: Formula
Relative lowering of vapour pressure ($\frac{P^\circ - P}{P^\circ}$) $= X_2$ (mole fraction of solute).
$X_2 = \frac{n_2}{n_1 + n_2}$
Step 2: Calculate Moles
$n_2$ (urea) $= \frac{3}{60} = 0.05 \text{ mol}$
$n_1$ (water) $= \frac{50}{18} = 2.77 \text{ mol}$
Step 3: Calculate $X_2$
$X_2 = \frac{0.05}{2.77 + 0.05} = \frac{0.05}{2.82} \approx 0.0177 \approx 0.018$
Final Answer: (C)