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

Answer

`a. Mw_(2)=22.71g mol^(-1)` `b. P^(@)=3.54kPa`

Explanation

`n_(2)=(30g)/(Mw_(2)),n_(1)=(90g)/(18g mol^(-1))=5 mol`
Use the relation,
`(p^(@)-p_(S))/(p_(S))=(n_(2))/(n_(1))implies(p^(@)-2.8)/(2.8)=(30//Mw_(2))/(5)`
`implies5xx((p^(@)-2.8)/(2.8))=30//Mw_(2) ….(i)`
After adding `18g H_(2)O( i.e., 1 mol of H_(2)O)`,
`n_(1)=6 mol`
`:. (P^(@)-2.9)/(2.9)=(30//Mw_(2))/(6)`
`implies6xx((p^(@)-p_(S))/(2.9))=30//Mw_(2) .....(ii)`
Equates `Eqs. (i) `and `(ii)`
`5((p^(@)-2.8)/(2.8))=6((p^(@)-2.9)/(2.9))`
Solve for `p^(@):`
`p^(@)=2.54kPa`
Substitute the value of `p^(@)` either in `Eq. (i) ` or in `Eq. (ii).`
`5xx((3.54-2.8)/(2.8))=(30)/(Mw_(2))`
Solve for `Mw_(2), `
`:. Mw_(2)=22.71g mol^(-1)`