Calculate vapour pressure of a `5%` (by weight ) solution of water
in glycerol (mol.wt.`92.1`) at `100^(@)C`, assuming Raoult's law to be
valid and neglecting the vapour pressure of glycerol.

Explanation

Since pure water boils at `100^(@)C` and b.p. is defined as that temperture at which the vapour pressure of the liquid is equal to atmospheric pressure , i.e. 760mm, the vapour pressure of pure water at `100^(@)C` is 760mm.
In this problem since glycerol `(92.1)` is the solute and water is the solvent and their weight per cents are `95%` and `5%` respectively, Raoult.s law cannot be applied as the solution is not at all dilute. But according to the question, we apply Raoult.s law
`(p^(0)-p)/(p^(0))=(n_(1))/(n_(1)+n_(2))`............[Eqn.1)
`{:("Here" p^(0)=760mm,,n_(1)="mole of glycerol"=(95)/(92.1)),(,,n_(2)="mole of water"=(5)/(18)):}}"supposing the weight of the solution as 100g"`
Substituting in Eqn. (1), we get `p=161.2mm`