The freezing point of equimolal aqueous solutions will be highest for which of the following?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C6H12O6
Depression of freezing point is a colligative property: it depends on the number of solute particles. For an electrolyte the van’t Hoff factor i counts the particles each formula unit gives:
\( \displaystyle \Delta T_f=i\,K_f\,m \)
The molality m is the same for all four, so the smallest i gives the smallest depression and the highest freezing point.
\( \ce{C6H5NH3Cl} \) — \( \ce{C6H5NH3Cl -> C6H5NH3^+ + Cl^-} \), so i = 2.
\( \ce{Ca(NO3)2} \) — \( \ce{Ca(NO3)2 -> Ca^2+ + 2NO3^-} \), so i = 3.
\( \ce{La(NO3)3} \) — \( \ce{La(NO3)3 -> La^3+ + 3NO3^-} \), so i = 4.
\( \ce{C6H12O6} \) (glucose) — a non-electrolyte; it does not dissociate, so i = 1.
Glucose has the lowest i, so its solution has the highest freezing point.
NCERT (2026-27 edition):
1.6 Colligative Properties and Determination of Molar Mass – “… There are many properties of solutions which are connected with this decrease of vapour pressure. These are: (1) relative lowering of vapour pressure of the solvent (2) depression of freezing point of the solvent (3) elevation of boiling point of the solvent and (4) osmotic pressure of the solution. All these properties depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.”
1.7 Abnormal Molar Masses – “In 1880 van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. … In case of association, value of i is less than unity while for dissociation it is greater than unity.”