A solution contaning 30 g of a non-volatile solute exactly in 90 g of water has a water has a vapour pressure of 2.8 K Pa at 298 K. Further 18 g of water is added to the solution and the new vapour pressure become 2.9 Pa at 298 K. Calculate.
Molecular mass of the solute.
Vapour presure of water of water at 2198 K.

Explanation

`(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)`
For I case: `(P^(@)-2.8)/(2.8) = (30 xx 18)/(m xx 90) = (6)/(m)` …(1)
For II case: `(P^(@)-2.9)/(2.9) = (30 xx 18)/(m xx 108) = (5)/(m)` …(2)
By eqs. (1) and (2),
`P^(@) = 3.53 kPa`
`m = 23 g mol^(-1)`
Note: Answers will be `3.4 kPa` and `34 g mol^(-1)` if `(P^(@)-P_(S))/(P^(@)) = (n)/(N)` is used which is only valid for dilute solutions.