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

Explanation

(i) Let, the molar mass of the solute be M g `mol^(-1)`
Now, the no. of moles of solvent (water),
`n_(1)=(90g)/(18g mol^(-1))=5 mol`
And, the no. of solute,
`n_(2)=(30g)/(M mol^(-1))=(30)/(M)mol`
`p_(1)=2.8` kPa
Applying the relation :
`(p_(1)^(0)p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))`
`therefore (p_(2)^(0)-2.8)/(p_(1)^(0)=((30)/(M))/(5+(30)/(M))`
`therefore 1-(2.8)/(p_(1)^(0))=((30)/(M))/((5M+30)/(M))`
`therefore 1-(2.8)/(p_(1)^(0))=(30)/(5M+30)`
`therefore (2.8)/(p_(1)^(0))=1-(30)/(5M+30)`
`therefore (28)/(p_(1)^(0))=(5M+30-30)/(5M+30)`
`therefore (2.8)/(p_(1)^(0))=(5M)/(5M+30)`
`therefore (p_(1)^(0))/(2.8)=(5M+30)/(5M) " "` ....(i)
After the addition of 18 g of water :
`n_(1)=(90+18)/(18)=6mol`
`p_(1)=2.9` kPa
Aganin, applying the relation :
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))`
`therefore (p_(1)^(0)-2.9)/(p_(1)^(0))=((30)/(M))/(6+(30)/(M))`
`therefore 1-(2.9)/(p_(1)^(0))=((30)/(M))/((6M+30)/(M))`
`therefore 1-(2.9)/(p_(1)^(0))=(30)/(6M+30)`
`therefore (2.9)/(p_(1)^(0))=1-(30)/(6M+30)`
`therefore (2.9)/(p_(1)^(0))=(6M+30-30)/(6M+30)`
`therefore (2.9)/(p_(1)^(0))=(6M)/(6M+30)`
`therefore (p_(1)^(0))/(2.9)=(6M+30)/(6M) " "` ....(ii)
Dividing equation (i) or (ii), we have :
`(2.9)/(2.8)=((5M+30)/(5M))/((6M+30)/(6M))`
`(2.9)/(2.8)xx(6M+30)/(6)=(5M+30)/(5)`
`2.9xx5xx(6M+30)=2.8xx6xx(5M+30)`
`87M+435=84M+504`
3 M = 69
M = 23
Therefore, the molar mass of the solute is 23 g `mol^(-1)`.
(ii) Putting the value of M in equation (i), we have :
`(p_(1)^(0))/(2.8)=(5xx23+30)/(5xx23)`
`therefore (p_(1)^(0))/(2.8)=(145)/(115)`
`therefore p_(1)^(0)=3.53` kPa
Hence, the vapour pressure of water at 298 K is 3.53 kPa.