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Depression in Freezing Point

1.1 g CoCl3...

Question

1.1 g $C o C l_{3} \cdot 6 N H_{3}$ (Molar mass 267.5) was dissolved in 100 g of water. The freezing point of solution was $- {0.306}^{o} C$ . How many moles of solute particles exist in solution for each mole of solute introduced? $(K_{f}$ for $H_{2} O = 1.86 K m o l^{- 1} k g)$ .

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Solution

The molality of the solution is $m = \frac{\frac{1.1 g}{267.5 g / m o l}}{0.1 k g} = 0.04112 m o l / k g$ .
The expression for the depression in the freezing point is $Δ T_{f} = i K_{f} m$
Substitute values in the above expression.
$0.306 = i \times 1.86 \times 0.04112$
$i = 4$
Thus moles of solute particles exist in solution for each mole of solute introduced.

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