Match the LIST-I with LIST-II.LIST-I(Expression)LIST-II(Law/Property)…

The correct answer is Option (3) → A-IV, B-III, C-II, D-I

All four expressions describe the vapour above a solution. Raoult’s law relates the partial vapour pressure of a component to its mole fraction in the solution; Dalton’s law adds the partial pressures; Henry’s law gives the solubility of a gas; and the relative lowering of vapour pressure follows from Raoult’s law for a solution of a non-volatile solute.

• A. \( p_a=p_a^{o}x_a \) — This matches IV. Raoult’s law. The partial vapour pressure of each volatile component is its pure vapour pressure multiplied by its mole fraction in the solution.

• B. \( p_{total}=p_1^{o}+(p_2^{o}-p_1^{o})\,x_2 \) — This matches III. Dalton’s law of partial pressure. The total pressure is the sum of the partial pressures; with Raoult’s law and \( x_1 = 1 - x_2 \):

      \( \displaystyle p_{total} = p_1^0 x_1 + p_2^0 x_2 = p_1^0 (1 - x_2) + p_2^0 x_2 = p_1^0 + (p_2^0 - p_1^0)\,x_2 \)

• C. \( p=K_H\,x \) — This matches II. Henry’s law. The partial pressure of a gas over a solution is proportional to its mole fraction in the solution; \( K_H \) is the Henry’s law constant.

• D. \( \frac{p_1^{o}-p_1}{p_1^{o}}=\frac{n_2}{n_1} \) — This matches I. Relative lowering in vapour pressure. With a non-volatile solute, \( p_1 = p_1^0 x_1 = p_1^0 (1 - x_2) \), so the lowering divided by \( p_1^0 \) equals the mole fraction of the solute; for a dilute solution \( n_2 \ll n_1 \), so \( x_2 \approx n_2/n_1 \):

      \( \displaystyle \frac{p_1^0 - p_1}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} \)

Note: The same/similar question has been asked in different years by NTA as per the details given below: