A solution containing `30 g` of a non-volatile solute exactly in `90 g` water has a vapour pressure of `2.8 kPa` at `298 K`. Further `18 g` of water is then added to solution, the new vapour pressure becomes `2.9 kPa` at `298 K`. Calculate:
(i) molecular mass of the solute,
(ii) vapour pressure of water at `298 K`.

Explanation

Let the molar mass of solute `=" Mg mol"^(-1)`
`therefore` Moles of solute present
`=(30 g)/( "M g mol"^(-1))=(30)/(M)` mol
Moles of solvent present, `(n_(1))=(90)/(18)=5` moles.
` therefore (P^(@)-P_(s))/(P^(@))=(n_(2))/(n_(1)+n_(2))`
`(P^(@)-2.8)/(P^(@))=(n_(2))/(n_(1)+n_(2))`
`(P^(@)-2.8)/(P^(@))=(30//M)/(5+30//M)`
`1-(2.8)/(P^(@))=(30)/((5M+30))`
`1-(30)/(5M+30)=(2.8)/(P^(@))`
`1-(6)/(M+6)=(2.8)/(P^(@))`
`(M+6-6)/(M+6)=(2.8)/(P^(@))`
`(M)/(M+6)=(2.8)/(P^(@))`
`(P^(@))/(2.8)=1+(6)/(M)`
After adding 18 g of water,
Moles of water becomes
`=(90+18)/(18)=(108)/(18)=6` moles
`therefore (P^(@)-P_(s))/(P^(@))=(30//M)/(6+30//M)`
`P_(s)` New vapour pressure =2.9 kPa
`(P^(@)-2.9)/(P^(@))=(30 M)/(M(6M+30))=(5)/(M+5)`
`1-(2.9)/(P^(@))=(5)/(M+5)`
`1-(5)/(M+5)=(2.9)/(P^(@))`
`(M+5-5)/(M+5)=(2.9)/(P^(@))`
` (P^(@))/(2.9)=(M+5)/(M) implies =1+(5)/(M)`
`(P^(@))/(2.9)=1+(5)/(M)`
Dividing equation (i) by (ii), we get,
`(2.9)/(2.8)=(1+6//M)/(1+5//M)`
`2.9(1+(5)/(M))=2.8(1+(6)/(M))`
`2.9+(2.9xx5)/(M)=2.8+(2.8xx6)/(M)`
`2.9+(14.5)/(M)=2.8+(16.8)/(M)`
`0.1=(16.8)/(M)-(14.5)/(M)=(2.3)/(M)`
`M=(2.3)/(0.1)`
`M=23 " g mol"^(-1)`
Putting M=23, in equation (i), we get,
`(P^(@))/(2.8)=1+(6)/(23)=(29)/(23)`
`P^(@)=(29)/(23)xx2.8=3.53 kPa`.