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Depression in Freezing Point

3.9g of benzo...

Question

3.9g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the Van't Hoff factor and predict the nature of solute (associated or dissociated).
[Given: Molar mass of benzoic acid =122 g $m o l^{-}, k_{f}$ for benzene=4.9 K kg $m o l^{- 1}$ ]

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Solution

Given : Mass of solute $(W_{B}) i . e, C_{6} H_{5} C O O H = 3.9 g$

Mass of solvent $(W_{A}) = 49 g = \frac{49}{1000} k g$

Molar mass of $C_{6} H_{5} C O O H (M_{S}) = 122 g / m o l$

$K_{f} = 4.9 K k g / m o l$

$Δ T_{f} = 1.62 K$
We know that the depression in freezing point is given by:

$Δ T_{f} = i \times K_{f} \times m$

$Δ T_{f} = i \times K_{f} \times \frac{W_{B}}{M_{S}} \times \frac{1}{W_{A} (K g)}$

$i = \frac{Δ T_{f} \times M_{S} \times W_{A} (K g)}{K_{f} \times W_{B}}$

$i = \frac{1.62 \times 122 \times 49}{4.9 \times 3.9 \times 1000}$

$i = 0.5067$

Since, $i < 1$ , hence solute benzoic acid $(C_{6} H_{5} C O O H)$ is associating.

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