Arrange the following aqueous solutions of equal mole fraction of sol…

The correct answer is Option (4) → C, D, A, B

Adding a non-volatile solute lowers the vapour pressure of water. By Raoult’s law the relative lowering equals the mole fraction of the solute particles. An electrolyte breaks into ions, so the effect is multiplied by the van’t Hoff factor i (the number of particles per formula unit):

    \( \displaystyle \frac{p^\circ-p}{p^\circ}=i\,x_2 \)

With the same mole fraction x₂ for all four solutes, the larger i, the larger the lowering and the lower the vapour pressure. Counting the particles:

• A. NaCl: NaCl → \( \ce{Na^{+}} \) + \( \ce{Cl^{-}} \): 1 + 1 = 2 ions, so i = 2.

• B. Urea: a non-electrolyte, does not dissociate: 1 particle, i = 1.

• C. \( \ce{Fe(NO3)3} \): \( \ce{Fe(NO3)3} \) → \( \ce{Fe^{3+}} \) + \( \ce{3NO3^{-}} \): 1 + 3 = 4 ions, so i = 4.

• D. \( \ce{CaCl2} \): \( \ce{CaCl2} \) → \( \ce{Ca^{2+}} \) + \( \ce{2Cl^{-}} \): 1 + 2 = 3 ions, so i = 3.

Lowering of vapour pressure: C (4) > D (3) > A (2) > B (1), so the vapour pressure itself is in the reverse order:

    \( \ce{Fe(NO3)3} \) < \( \ce{CaCl2} \) < NaCl < urea

Shortcut: \( \ce{Fe(NO3)3} \) gives the most particles, so its solution has the lowest vapour pressure. Only Option (4) starts with C.