Question

In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl\(_2\) solution is?

• A. The same
• B. About twice
• C. About three times
• D. About six

Hint:

The van’t Hoff factor (\(i\)) equals the number of particles formed after dissociation and determines the magnitude of colligative properties.

Answer 1

The Correct Option is C

Step 1: Depression in freezing point is a colligative property and depends on the number of solute particles present in the solution.
Step 2: Glucose is a non-electrolyte and does not dissociate in solution, so one molecule produces one particle.
Step 3: Magnesium chloride (\(\text{MgCl}_2\)) is an electrolyte and dissociates as: \[ \text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- \] Thus, one formula unit produces three particles.
Step 4: Hence, for the same molar concentration, the depression in freezing point of \(\text{MgCl}_2\) solution is approximately three times that of glucose solution.