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Depression in Freezing Point

15.0 g of an ...

Question

15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at $- {0.34}^{o} C$ . What is the molar mass of this material? ( $K_{f}$ for water=1.86 $K k g m o l^{- 1}$ )

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Solution

(b) Given mass of solute $= w_{b} = 15.0 g$
Molar mass of solute $= M_{b} = ?$
Mass of water $= w_{a} = 450 g$
Freezing point of water $= 0 {^A}^{o} C = 273 K$
Freezing point of solution $= - 0.34 {^A}^{o} K$
$= (- 0.34 + 273) K$
Depression in freezing point $=\in" T_{f} = 273 - (- 0.34 + 273) = 0.34 K$
$K_{f}$ for water = 1.86 K Kg $m o l^{- 1}$
Since, $Δ T_{f} = K_{f} \times m = K_{f} \frac{w_{b} \times 1000}{M_{b} \times w_{a}}$
$M_{b} = \frac{K_{f} \times w_{b} \times 1000}{w_{a} Δ T_{f}}$
$M_{b} = \frac{1.86 \times 15 \times 1000}{450 \times 0.34}$
$= 182.35 g / m o l$

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