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

Explanation

Calculation of molecular mass of the solute.
In the first case.
`"No. of moles of solute"(n_(B))=("Mass of solute")/("Molecular mass")=((30g))/((M g mol^(-1)))=30/M mol`
`"No. of moles of solute"(n_(A))=("Mass of solute")/("Molecular mass")=((90g))/((18 g mol^(-1)))=5 mol`
`"Mole fraction of water"(x_(A))=n_(A)/(n_(A)+n_(B))=((5mol))/((5mol+30/Mmol))=5/(5+30/Mmol)=(5M)/((5M+30))=M/((M+6))`
"Vapour pressure of solution"`(P_(A))`=2.8 kPa
"According to Raoult's Law, `P_(A)=P_(A)^(@)x_(A)`
`(2.8 kPa)=P_(A)^(@)M/((M+6))`
In the second case :
` "No. of moles of of solute" (n_(B))=30?M mol `
`"No. of moles of water" (n_(A))=("Mas of water")/("Molecular mass")=((108g))/((18 g mol^(-1)))=6 mol`
`"Mole fractio of water" (x_(A))=n_(A)/(n_(B)+n_(B))=((6mol))/((6mol+30/Mmol))=(6M)/((6M+30))+M/((M+5))`
"Vapour pressure of solution"`(P_(A))=2.9 kPa`
`"According ro Raoult's Law", P_(A)=P_(A)^(@)x_(A)`
` (2.9 kPa)=P_(A)^(@)M/((M+5))`
` "Dividing eqn. (i) by eqn. (ii)," `
`((2.8 kPa))/((2.9kPa))=((M+5))/((M+6))`
`((M+5))/((M+6))=(2.8)/(2.9)or 0.9655`
`(M+5)=0.9655 xx6+0.9655 M`
0.0345M =0.793
` M=(0.793)/(0.0345)=23 or 23 u`
Calculation of vapour pressure of water
According to Raoult's Law , `P_(A)=P_(A)^(@)x_(A)or (2.8 kPa)=P_(A)^(@)5/((5+30/M))`
`(2.8 kPa)=P_(A)^(@)5/((5+30/23.0))=P_(A)^(@)5/((5+1.30))=P_(A)^(@)5/6.30`
`P_(A)^(@)=((2.8 kPa)xx6.30)/5=3.53 kPa`