(a) State Raoult’s law for a solution containing volatile components.

How does Raoult’s law become a special case of Henry’s law?

(b) 1·00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0·40 K. Find the molar mass of the solute. (K_f for benzene = 5·12 K kg mol^-1) 

(a) Let p₁, p₂ = Partial vapour pressure of two volatile components 1 and 2 of a mixture 

= Vapour pressure of pure components 1 and 2

x₁, x₂ = Mole fractions of the components 1 and 2

p_total = Total vapour pressure of the mixture then Raoult’s law can be stated as: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

That is, for component 1,

p₁      x₁

And, p₁ = x₁

For component 2,

P₂ = x₂

According to Dalton’s law of partial pressures,

 The plot of vapour pressure and mole fraction of an ideal solution at constant temperature is shown below.

Raoult’s Law as a Special Case of Henry’s Law

According to Raoult’s law, the vapour pressure of a volatile component in a given solution is p₁ = x₁

According to Henry’s law, the partial vapour pressure of a gas (the component is so volatile that it exists as gas) in a liquid is

p = K_H x

It can be observed that in both the equations, the partial vapour pressure of the volatile component varies directly with its mole fraction. Only the proportionality constants K_H and are different. Thus, Raoult’s law becomes a special case of Henry’s law in which K_H is equal to .

(b) Given: w₂ = 1g (weight of solute)

w₁ = 50 g (weight of solvent)

T_f = 0.40 K

k_f = 5.12 K Kg mol^-1

M₂ =? (Molar mass of solute)

Using the equation,

T_f = K_fm (where m is molality)

0.40 = 5.12 x m