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3.9 g 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 mol^{−1}, K_{f} for benzene = 4.9 K kg mol^{−1})

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#### Solution

We know that the depression in freezing point is given by

`DeltaT_`

Here,

van't Hoff factor = i

Depression in freezing point, ΔT_{f}=1.62 K

K_{f} for benzene=4.9 K kg mol^{−1}

Mass of benzoic acid, w_{s}=3.9 g

Mass of benzene, W=49 g

Molar mass of benzoic acid, M_{s}=122 g mol^{−1}

Substituting the values, we get

`1.62 = (ixx4.9xx(3.9xx1000))/(122xx49)`

`rArri=(1.62xx122xx49)/(4.9xx3.9xx1000)`

**= 0.51**

As the value of i < 1, benzoic acid is an associated solute.

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