A solution containing 30g of non-volatile solute exactly in 90g of water has a vapour pressure of `2.8" kP"_(a)` at 298 K. Further 18g of water is then added to the solution and the new vapur pressure becomes `2.9" kP"_(a)` at 298 K. Calculate (i) The moar mass of the solute and (ii) Vapour pressure of water at 298 K.

Explanation

`w_(B) = 30 g , w_(A) =90 g, p_(A) = 2.8 kPa`
According to Raoult.s law,
`(p_(A)^(@) - p_(A))/(p_(A)^(@)) = x_(B) = ((w_(B))/(M_(B)))/((w_(A))/(M_(A)) + (w_(B))/(M_(B))) = (w_(B) xx M_(B))/(M_(B) xx w_(B))`
`therefore " " (p_(A)^(@) - 2.8)/(p_(A)^(@)) = (30)/(M_(B)) xx (18)/(90)`
`1- (2.8)/(p_(A)^(@)) = 1 - (6)/(M_(B))`
or `" " (2.8)/(p_(A)^(@)) = (M_(B) - 6)/(M_(B)) " "....(i)`
Similarly, `w_(B) = 30 g , w_(A) = 90 + 18 = 108 g , p_(A) = 2.9 kPa`
`therefore " " (p_(A)^(@) - 2.9)/(p_(A)^(@)) = (30)/(M_(B)) xx (18)/(108)`
` 1- (2.9)/(p_(A)^(@)) = (5)/(M_(B))`
or `" " (2.9)/(p_(A)^(@)) = 1 - (5)/(M_(B))`
or `" " (2.9)/(p_(A)^(@)) = (M_(B) - 5)/(M_(B))" "........(ii)`
Dividing eq.(i) by . (ii) we get .
`(2.8)/(2.9) = (M_(B) - 6)/(M_(B) - 5)`
`2.8 M_(B) - 14 = 2.9 - 17.4`
` 2.9 M_(B) - 2.8 M_(B) = 17.4 - 14`
`0.1 M_(B) = 3.4`
`therefore " " M_(B) = 34 g mol^(-1)`
Subsituting the value of `M_(B)` in eqn. (i) we get
`(2.8)/(p_(A)^(@)) = (34-6)/(34) = (28)/(34)`
`therefore " " 28p_(A)^(@) = 2.8 xx 34`
`therefore " "p_(A)^(@) = 3.4 kPa`