What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol$$^{-1}$$

We need to find the weight of glucose to dissolve in 100 g of water to lower vapour pressure by 0.20 mm Hg.

$$P^\circ = 54.2$$ mm Hg, $$\Delta P = 0.20$$ mm Hg, molar mass of glucose = 180 g/mol, molar mass of water = 18 g/mol.

Apply Raoult's law for dilute solutions.

For a dilute solution, the relative lowering of vapour pressure is:

$$ \frac{\Delta P}{P^\circ} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} $$

(For dilute solutions, $$n_{\text{solute}} + n_{\text{solvent}} \approx n_{\text{solvent}}$$, so this simplification is valid.)

Express moles in terms of given quantities.

$$n_{\text{solute}} = \frac{w}{180}$$ (where $$w$$ is the weight of glucose in grams)

$$n_{\text{solvent}} = \frac{100}{18}$$

Substitute and solve.

$$ \frac{0.20}{54.2} = \frac{w/180}{100/18} = \frac{w}{180} \times \frac{18}{100} = \frac{w}{1000} $$

$$ w = \frac{0.20 \times 1000}{54.2} = \frac{200}{54.2} = 3.69\;\text{g} $$

The weight of glucose required is 3.69 g.

The correct answer is Option 2: 3.69 g.