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^@)=17.51 mm`
Lowering of vapour pressure`(P^(@)-P_(s))=0.0614 mm`
a. Relative lowering of vapour pressure
`(P^(@)-P_(s))/P^(@)=0.0614/17.51 =0.00351`
b. Vapour pressure of the solution
`P_(s)=P^(@)-(P^(@)-P_(s))=17.51-0.0614=17.4486 mm`
c. To calculate the mole fraction of water
By Raoult's law `=(P^(@)-P_(s))/P^(@)`
`n_(2)/n_(1)+n_(2)`
`=chi_(2)`,mole fraction of solute
i.e., `chi_(2)=(P^(@)-P_(s))/P^(@)=0.00351`
`:.`Mole fraction of solvent (water),
`chi_(1)=1-chi_(2)=1-0.00351=0.99649`