Two different non-volatile solutes are dissolved together in a volatile solvent. The lowering of the vapour pressure of the solvent in this solution depends on:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The sum of the mole fractions of the two solutes
By Raoult’s law, the lowering of vapour pressure depends only on the concentration of solute particles, not on their identity. For one solute it equals the mole fraction of the solute multiplied by the vapour pressure of the pure solvent:
\( \displaystyle \Delta p_1 = x_2\, p_1^0 \)
With two solutes, the particles of both take the place of solvent particles, so their mole fractions add up:
\( \displaystyle \Delta p_1 = (x_2 + x_3)\, p_1^0 \)
Why the other options are incorrect
Option (2) The mole fraction of the solute having the higher molar mass — Incorrect. The particles of both solutes lower the vapour pressure; molar mass does not pick out one of them.
Option (3) The sum of the masses of the two solutes — Incorrect. Equal masses of two solutes with different molar masses contain different numbers of moles. The lowering follows the number of moles of particles, not the mass.
Option (4) The product of the mole fractions of the two solutes — Incorrect. The particles of the two solutes add to one another, so the mole fractions are added, not multiplied.
NCERT (2026-27 edition):
1.6.1 Relative Lowering of Vapour Pressure – “Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity. … Knowing that x2 = 1 – x1, equation (1.23) reduces to Δp1 = x2 p10 (1.24) In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.”