Question

# (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. (Kf for benzene = 5·12 K kg mol-1)

(a) Let p1p2 = Partial vapour pressure of two volatile components 1 and 2 of a mixture

= Vapour pressure of pure components 1 and 2

x1x= Mole fractions of the components 1 and 2

ptotal = 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 = KH 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 KH and are different. Thus, Raoult’s law becomes a special case of Henry’s law in which KH is equal to .

(b) Given: w2 = 1g (weight of solute)

w1 = 50 g (weight of solvent)

Tf = 0.40 K

kf = 5.12 K Kg mol-1

M2 =? (Molar mass of solute)

Using the equation,

Tf = Kfm (where m is molality)

0.40 = 5.12 x m