The vapour pressure of pure solvent is 0.8 mm of Hg at a particular temperature. On addition of a no — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

According to Raoult's law, the lowering in vapour pressure is written as, 0 2 0 P P X P We are provided with the following data: Vapour pressure of the pure solvent is 0P 0.8 mm of Hg Vapour pressure of solvent on the addition of solute is P 0.6 mm of Hg Let's substitute the values in the equation. We have, 2 2 0.8 0.6 X 0.8 X 0.25 >> Thus, (a) is the correct option.