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Van't Hoff Factor

The freezing ...

Question

The freezing point of a solution containing 5.85 g of $N a C l$ in 100g of water is $- {3.348}^{o}$ C. Calculate van't Hoff factor 'i' for this solution. What will be the experimental molecular weight of NaCI?
( $K_{f}$ for water = $1.86 K k g m o l^{- 1}$ , at. wt. Na = 23, Cl = 35.5)

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Solution

$w_{N a C l}$ =5.85 g; $w_{H_{2} O}$ = 100 g; $Δ T_{f}$ = 3.348 K; $K_{f}$ =1.86 Kkg/mol
As, $Δ T_{f} = i K_{f} m m = \frac{w_{N a C l}}{M_{N a C l}} \times \frac{1000}{w_{H_{2} O}} m = \frac{5.85}{58.5} \times \frac{1000}{100} = 1 T h e r e f o r e, i = \frac{3.348}{1.86} = 1.8 A l s o, i = \frac{N o r m a l m o l e c u l a r w t}{A b n o r m a l m o l e c u l a r w t} 1.8 = \frac{58.5}{M} ∴ M = 32.5 g / m o l$
Vant Hoff factor is $1.8$ and experimental molecular weight of $N a C l$ is 32.5 g/mol.

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