\(0.1 M\, \ NaCl\) solution has higher freezing point than \(0.1 M \, \ MgCl_2\) solution in water because:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → van’t Hoff factor for \(MgCl_2\) is more than \(NaCl\)
Depression of freezing point is a colligative property: it depends on the number of solute particles, not on their nature. For an electrolyte the van’t Hoff factor i enters the equation:
\( \displaystyle \Delta T_f = i\,K_f\,m \)
\( \displaystyle \ce{NaCl -> Na+ + Cl-}\qquad i = 2 \)
\( \displaystyle \ce{MgCl2 -> Mg^2+ + 2Cl-}\qquad i = 3 \)
At the same concentration, \( \ce{MgCl2} \) gives more particles, so its ΔTf is larger (about 3/2 times) and its freezing point is lower. The NaCl solution therefore has the higher freezing point, because the van’t Hoff factor of MgCl2 is more than that of NaCl.
Why the other options are incorrect
Option (1) \(MgCl_2\) is less ionic than \(NaCl\) — Incorrect. The comparison is decided by the number of ions per formula unit, not by ionic character; both salts are strong electrolytes.
Option (2) van’t Hoff factor (i) for \(NaCl\) is more than \(MgCl_2\) — Incorrect. i is 2 for NaCl and 3 for \( \ce{MgCl2} \), so this is the reverse of the truth.
Option (4) \(NaCl\) is less ionic than \(MgCl_2\) — Incorrect. Again ionic character is not the deciding factor; the extra \( \ce{Cl^{-}} \) ion from \( \ce{MgCl2} \) is.
NCERT (2026-27 edition):
Abnormal Molar Masses – “… Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: Relative lowering of vapour pressure of solvent, (p1° – p1)/p1° = i.n2/n1; Elevation of Boiling point, ΔTb = i Kb m; Depression of Freezing point, ΔTf = i Kf m; Osmotic pressure of solution, Π = i n2 R T / V.”