Solutions

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. (K_{f} for benzene = 5·12 K kg mol^{-1})

Answer

(a) Let *p*_{1}, *p*_{2} = Partial vapour pressure of two volatile components 1 and 2 of a mixture

= Vapour pressure of pure components 1 and 2

*x*_{1}, *x*_{2 }= 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,

*p*_{1 } *x*_{1}

And, p_{1} = x_{1}

For component 2,

P_{2} = x_{2}

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 p_{1} = x_{1}

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: w_{2} = 1g (weight of solute)

w_{1} = 50 g (weight of solvent)

T_{f} = 0.40 K

k_{f} = 5.12 K Kg mol^{-1}

M_{2} =? (Molar mass of solute)

Using the equation,

T_{f} = K_{f}m (where m is molality)

0.40 = 5.12 x m