The vapour pressure of water at `293 K` is `17.51 mm`. The lowering of vapour pressure of sugar is `0.0614 mm`. Calculate:
a. The relative lowering of vapour pressure
b.The vapour pressure of the solution
c. The mole fraction of water

Explanation

Here, we are given that Vapour pressure of water `(p^(@))=17.51mm`
Lowering of vapour pressure `(p^(@)-p_(S))=0.0614mm`
(i)`"Relative lowering of vapour pressure "=(p^(@)-p_(s))/(p^(@))=("0.0614 mm")/("17.51 mm")="0.00351"`
(ii) Vapour pressure of the solution `(p_(s))`
`p^(@)-p_(s)="0.0614 mm, "p^(@)="17.51 m"`
`therefore" 17.51mm "-p_(s)="0.0614 mm or "p_(s)="17.51 mm "-0.0614 mm = 17.4468 mm"`
(iii) To calculate mole fraction of water :
By Raoult's law `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))=x_(2),` mole fraction of solution, i.e., `x_(2)=(p^(@)-p_(s))/(p^(@))=0.00351`
`therefore" Mole fraction of solvent (water )," x_(1)=1-x_(2)=1-0.00351=0.99649`