Question

When \(10^{-3} \, M\) solution of glucose in water freezes at -0.0186°C, then at what temperature \(10^{-3} \, M\) solution of NaCl will freeze?

• A. \(0^\circ C\)
• B. \(0.186^\circ C\)
• C. \(-0.186^\circ C\)
• D. \(-0.0372^\circ C\)

Hint:

The freezing point depression is proportional to the molality and van't Hoff factor. Dissociation increases the freezing point depression.

Answer 1

The Correct Option is D

We use the freezing point depression formula: \[ \Delta T_f = i \cdot K_f \cdot m \] where: - \( \Delta T_f \) is the change in freezing point - \( i \) is the van't Hoff factor (number of particles) - \( K_f \) is the freezing point depression constant - \( m \) is the molality of the solution. For glucose, \(i = 1\) (since it does not dissociate), and for NaCl, \(i = 2\) (since it dissociates into two ions). Since the freezing point depression of glucose is given, we can calculate the depression for NaCl. For glucose: \[ \Delta T_f = 0.0186^\circ C \] For NaCl, using the same molarity: \[ \Delta T_f(\text{NaCl}) = 2 \times 0.0186^\circ C = 0.0372^\circ C \] Thus, the final temperature for NaCl solution will be: \[ \boxed{-0.0372^\circ C} \]