Statement I: When a liquid solute that is more volatile than the solvent is added to the solvent, the vapour pressure of the solution can become higher than that of the pure solvent, that is, ps > p°.
Statement II: In a solution of a more volatile liquid solute in a liquid solvent, only the solute forms vapour and the solvent does not.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Statement I is correct but Statement II is incorrect.
When both liquids are volatile, each one evaporates and gives its own partial vapour pressure. The total vapour pressure is their sum and, for an ideal solution, lies between the vapour pressures of the two pure liquids:
\( \displaystyle p_{total}=p_1+p_2=p_1^0+(p_2^0-p_1^0)\,x_2 \)
Statement I: When a liquid solute that is more volatile than the solvent is added to the solvent, the vapour pressure of the solution can become higher than that of the pure solvent, that is, ps > p°. — True. Take the solvent as component 1 and the more volatile solute as component 2, so p20 > p10. Then (p20 − p10)x2 is positive and ptotal rises above p10, the vapour pressure of the pure solvent.
Statement II: In a solution of a more volatile liquid solute in a liquid solvent, only the solute forms vapour and the solvent does not. — False. In a closed vessel both components evaporate, so the vapour contains the solvent as well as the solute.
NCERT (2026-27 edition):
1.4.1 Vapour Pressure of Liquid-Liquid Solutions – “Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase.”
“… The minimum value of ptotal is p10 and the maximum value is p20, assuming that component 1 is less volatile than component 2, i.e., p10 < p20.”