freezing point depression

Question

A solution of urea (molar mass = 60 g/mol) in water freezes at −0.372°C. If Kf for water is 1.86 K kg mol⁻¹, the molality of the solution is:

RankGuru · Accepted Answer

The depression of freezing point is given by:
$$\Delta T_f = K_f 	imes m$$
- **Given**:
  - Freezing point of solution $= -0.372^\circ C$.
  - Therefore, $\Delta T_f = 0 - (-0.372) = 0.372 	ext{ K}$.
  - $K_f = 1.86 	ext{ K kg mol}^{-1}$.
- **Calculation**:
  - $m = \frac{\Delta T_f}{K_f} = \frac{0.372}{1.86} = 0.2 	ext{ m}$.

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More freezing point depression Practice Questions

0.100 kg water contains 0.010 mol glucose (i=1). Kf(water)=1.86 K kg mol−¹. ΔTf is: