Calculate the lowering of vapour pressure caused by the addition of 50 g of sucrose (mol mass = $342\, g\, mol^{-1}$) to 750 g of water, if the vapour pressure of pure water at 25°C is 23.8 mm Hg.

0.083 mm Hg

The correct answer is Option (1) → 0.083 mm Hg

To find the lowering of vapour pressure, use Raoult’s Law:

$\Delta p = p^0 \cdot x_{\text{solute}}$

Step 1: Moles of sucrose

$\text{Moles} = \frac{50}{342} = 0.146 \text{ mol}$

Step 2: Moles of water

$\text{Moles of water} = \frac{750}{18} = 41.67 \text{ mol}$

Step 3: Mole fraction of solute

$x_{\text{solute}} = \frac{0.146}{41.67 + 0.146} \approx 0.0035$

Step 4: Lowering of vapour pressure

$\Delta p = 23.8 \times 0.0035 \approx 0.083 \text{ mm Hg}$

✔ Correct Answer: 0.083 mm Hg