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Relative Lowering of Vapour Pressure

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Question

Calculate the freezing point depression expected for $0.0711$ m aqueous solution of $N a_{2} S O_{4}$ . If this solution actually freezes at $- {0.320}^{o}$ C, what would be the value of van't Hoff factor? ( $K_{f}$ for water is $1.86$ K kg $m o l^{- 1}$ ).

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Solution

Depression in freezing point of a solvent upon aaddition of solute is given as:
$Δ T_{f} = m . K_{f}$
m=molality of the solution=0.0711 kg/mol
$K_{f}$ =cryoscopic constant=1.86 K kg/mol
$Δ T_{f} = 0.0711 \times 1.86 = 0.132$ K
thus freezing point of water is
$T_{s o l v e n t} = 273 - 0.132 = 272.868 K = - {0.132}^{0} C$
observed $Δ T_{f} = - {0.320}^{0} C$
vant hoff factor, i=observed freezing point/calculated freezing point
$∴ i = 0.32 / 0.132 = 2.42$

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