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

Solution

Chemistry ma’am · 4 steps

Relative lowering equals the solute's mole fraction, and in a dilute solution that is just n₂/n₁.

1. Raoult's law. For a non-volatile solute like glucose, the relative lowering of vapour pressure equals the mole fraction of the solute. In a dilute solution, n₂ is tiny next to n₁, so the mole fraction is close to n₂/n₁.

   (Δ P)/P° = n₂/n₁ = (w/180)/(100/18)

2. Put in numbers. The left side is 0.20/54.2 = 3.69 × 10⁻³. Water gives n₁ = 100/18 = 5.56 mol, so n₂ = 3.69 × 10⁻³ × 5.56 = 0.0205 mol.
3. Convert to mass. w = 0.0205 × 180 = 3.69 g. The shortcut works the same way: w = 0.20/54.2 × 100/18 × 180 = 0.20/54.2 × 1000 = 3.69 g.
4. So. About 3.69 g of glucose is needed, which is option (a).

The idea to keep. Relative lowering of vapour pressure equals the solute's mole fraction. When the question says dilute, use n₂/n₁ and you can solve for the solute mass in one line.

Answer from the NTA final answer key. Solution by PrepWiser.