Question

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)

Answer 1

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.