A solution containing 30g of a non-volatile non-electrolyte solute exactly in 90g water has a vapour pressure of `2.8 kPa` at 298K. Further, 18g of water is then added to solution, the new vapour pressure becomes `2.9kPa` at 298K. The solutions obey Raoult's law and are not dilute, molar mass of solute is

Explanation

Suppose the molar mass of the solute `="M g mol"^(-1)`
`n_(2)" (solute)"=(30)/(M)"moles, "n_(1) ("solvent, "H_(2)O)=("90 g")/("18 g mol"^(-1))="5 moles"`
`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2)), i.e., (p^(@)-2.8)/(p^(@))=(30//M)/(5+30//M)or1-(2.8)/(p^(@))=(30//M)/(5+30//M)`
`"or "(2.8)/(p^(@))=1-(30//M)/(5+30//M)=(5+30//M-30//M)/(5+30M)=(5)/(5+30//M)or (p^(@))/(2.8)=(5+30//M)/(5)=1+(6)/(M)" ...(i)"`
After adding 18 g of water, `n(H_(2)O)`, i.e., `n+(1)=6` moles
`therefore" "(p^(@)-2.9)/(p^(@))=(30//M)/(6+30//M)or 1-(2.9)/(p^(@))=(30//M)/(6+30//M)`
`"or "(2.9)/(p^(@))=1-(30//M)/(6+30//M)=(6+30//M-0//M)/(6+30//M)=(6)/(6+30//M)" or "(p^(@))/(2.9)=(6+30//M)/(6)=1+(5)/(M)" ...(ii)"`
Dividing eqn. (i) by eqn. (ii), we get `(2.9)/(2.8)=(1+6//M)/(1+5//M) " or "2.9(1+(5)/(M))=2.8(1+(6)/(M))`
`"or "2.9+(14.5)/(M)=2.8+(16.8)/(M) " or "(2.3)/(M)=0.1 " or "M = 23u`
(ii) Putting M = 23 in eqn. (i) we get `(p^(@))/(2.8)=1+(6)/(23)=(29)/(23)" or "p^(@)=(29)/(23)xx2.8="3.53 kPa."`