At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of the pure solvent is 0.80 atm. The lowering of vapour pressure is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.20 atm
By Raoult’s law, a non-volatile solute lowers the vapour pressure of the solvent, and the lowering of vapour pressure equals the mole fraction of the solute times the vapour pressure of the pure solvent:
\( \displaystyle \Delta p_1=p_1^0-p_1=x_2\,p_1^0 \)
Here x2 = 0.25 and p10 = 0.80 atm:
\( \displaystyle \Delta p_1=0.25\times 0.80\,\mathrm{atm} \)
\( \displaystyle \Delta p_1=0.20\,\mathrm{atm} \)
The lowering of vapour pressure is 0.20 atm.
NCERT (2026-27 edition):
1.6.1 Relative Lowering of Vapour Pressure – “We have learnt in Section 1.4.3 that the vapour pressure of a solvent in solution is less than that of the pure solvent. Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity. … The reduction in the vapour pressure of solvent (Δp1) is given as:
\( \displaystyle \Delta p_1=p_1^0-p_1=p_1^0-p_1^0x_1=p_1^0(1-x_1) \qquad (1.23) \)
Knowing that x2 = 1 – x1, equation (1.23) reduces to
\( \displaystyle \Delta p_1=x_2\,p_1^0 \qquad (1.24) \)”