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.

Calculate vapour pressure of a 5% (by weight ) solution of water in gl

1.	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.
Answer» Solution :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`