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

Calculation of molar mass of solute
Case - I :
Number of moles of solute `(n_(s))=("Weight")/("GMw")=(30)/(M)`
Number of moles of solvent `(n_(0))` (water) `=(90)/(18)=5`
Mole fraction of water `X_(0)=(n_(0))/(n_(0)+n_(s))=(5)/(5+(30)/(M))=(M)/(6+M)`
Vapour pressure of `1^("st")` solution `(P_(A))=2.8" kP"_(a)`
`P_(A)=P_(A)^(0)xx X_(A)`
`2.8=P_(A)^(0)xx(M)/(6+M)" "...(1)`
Case - II
No. of moles of solute `(n_(s))=(30)/(M)`
No. of moles of water `(n_(0))=(108)/(18)=6`
Mole fraction of water `(X_(0))=(n_(0))/(n_(0)+n_(s))=(6)/(6+(30)/(M))=(M)/(5+M)`
Vapour pressure of solution `P_(A)=2.9" kP"_(a)`
`P_(A)=P_(A)^(0)xx X_(A)`
`2.9=P_(A)^(0)xx("M")/(5+M)" "...(2)`
Dividing equation (1) by equation (2) `(2.8)/(2.9)=(5+M)/(6+M)`
`0.9655=(5+M)/(6+M)`
`0.0345" M"=0.793`
`M=(0.793)/(0.0345)=23" gms"//"mole."`
Calculation of vapour pressure of water :
AC Cording to Raoult's law
`P_(A)=P_(A)^(0)X_(A)`
`2.8=P_(A)^(0)xx("M")/(6+"M")`
M = 23
`2.8=P_(A)^(0)xx(23)/(6+23)`
`P_(A)^(0)=(2.8xx29)/(23)=(81.2)/(23)=3.53" k.p"_(a)`