An ideal solution is one in which the attraction between components of the solution is the same as the interaction between the molecules of each component. Heat is neither absorbed nor evolved during the formation of an ideal solution and the volume of the solution is equal to the sum of the volumes of the component liquids. Raoult’s law states that the partial vapour pressure of each component (say liquid A) of the solution is directly proportional to its mole fraction in the solution. The total vapour pressure over the solution is equal to the sum of the partial vapour pressures of the individual components.
A non-ideal solution is that solution (i) which does not obey Raoult’s law (ii) for which \(\Delta V_{mix}\) is not zero and (iii) for which \(\Delta H_{mix}\) is not zero. In non-ideal solutions, the solute–solvent interactions are weaker or stronger than the solute–solute and solvent–solvent interactions. The non-ideal solutions in which solute–solvent interactions are weaker than the solute–solute and solvent–solvent interactions show positive deviation from Raoult’s law, and those in which they are stronger show negative deviation.
Chloroform (CHCl3) and dichloromethane (CH2Cl2) form an ideal solution. At 298 K, the vapour pressures of pure CHCl3 and pure CH2Cl2 are 200 mm Hg and 415 mm Hg, respectively. What must be the mole fraction of CHCl3 in the liquid so that the vapour in equilibrium with it at 298 K contains equal mole fractions of the two components?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.675
By Dalton’s law of partial pressures, the mole fraction of a component in the vapour is \( y_i = p_i/p_{total} \). Equal mole fractions in the vapour therefore need equal partial pressures.
By Raoult’s law each partial pressure is \( x_i\,p_i^0 \). Let x be the mole fraction of CHCl3 in the liquid; then the mole fraction of CH2Cl2 is (1 − x).
\( \displaystyle p_{\ce{CHCl3}} = p_{\ce{CH2Cl2}} \)
\( \displaystyle x \times 200 = (1 - x) \times 415 \)
\( \displaystyle 615\,x = 415 \)
\( \displaystyle x = \frac{415}{615} = 0.675 \)
The liquid must hold more of CHCl3, the less volatile component, to make up for its lower vapour pressure.
NCERT (2026-27 edition):
1.4.1 Vapour Pressure of Liquid-Liquid Solutions – “…for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.”
\( \displaystyle p_1 = p_1^0\,x_1 \qquad (1.12) \)
“The composition of vapour phase in equilibrium with the solution is determined by the partial pressures of the components. If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures: p1 = y1 ptotal, p2 = y2 ptotal. In general pi = yi ptotal”