Question

At T (K), x g of a non-volatile solid (molar mass 78 g mol$^{-1}$) when added to 0.5 kg water, lowered its freezing point by 1.0°C. What is x (in g)? (K$_f$ of water at T(K) = 1.86 K kg mol$^{-1}$)

• A. 10.48
• B. 20.96
• C. 41.92
• D. 5.24

Hint:

Use $\Delta T_f = K_f \cdot \dfrac{1000 \cdot w}{M \cdot W}$ where $w$ is solute weight, $W$ is solvent mass.

Answer 1

The Correct Option is B

We use the formula for depression in freezing point: $\Delta T_f = K_f \cdot \dfrac{w \cdot 1000}{M \cdot W}$ Given: $\Delta T_f = 1^\circ C$, $K_f = 1.86$, $M = 78$ g/mol, $W = 500$ g (0.5 kg) Substitute: $1 = 1.86 \cdot \dfrac{x \cdot 1000}{78 \cdot 500}$ $\Rightarrow x = \dfrac{1 \cdot 78 \cdot 500}{1.86 \cdot 1000} = 20.96$ g