Question

Distilled water boils at 373.15 K and freezes at 273.15 K. A solution of glucose in distilled water boils at 373.202 K. What is the freezing point (in K) of the same solution? (For water, \( K_b = 0.52 \) kg mol\(^{-1}\), \( K_f = 1.86 \) kg mol\(^{-1}\))

• A. 273.15
• B. 273.0
• C. 272.964
• D. 273.336

Hint:

Boiling point elevation and freezing point depression depend on molality and colligative properties.

Answer 1

The Correct Option is C

- The boiling point elevation is: \[ \Delta T_b = i K_b m \] \[ 0.052 = (1)(0.52) m \] \[ m = 0.1 \text{ mol/kg} \] - The freezing point depression is: \[ \Delta T_f = i K_f m \] \[ \Delta T_f = (1)(1.86)(0.1) = 0.186 \] \[ T_f = 273.15 - 0.186 = 272.964 K \]