An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following.
Lowering in vapour pressure at 298 K is

Explanation

Given Freezing point of solution `=272.4K`
Molal elevation constant of water `=273K`
Molal elevation constant of water `=0.512K kg "mol"^(-1)`
Cryoscopic constant of water `=1.86K`
Vapour pressure of pure solvent `=23.757mm Hg`.
Asked (i) Molality of solution `=?`
(ii) Boiling point of solution
(iii) Lowering of vapour pressure of water
`ul(" Calculation of Molality")`
Formulae: `DeltaT_(f)=K_(f)xxm`
Explanation: `DeltaT_(f)=`Depression in freezing point
`K_(f)=` Cryoscopic constant
`m=` Molality
Substitution `&` Calculation
`DeltaT_(f)=273-272.4=0.6K rArr 0.6K=1.86xxm`
`m=0.322 "mol"//kg`
Molality of solution is `0.322m`
`ul("Calculation of Boiling point of solution")`
Formulae: `DeltaT_(b)=K_(b)xxm`
Explanation: `DeltaT_(b)=` Elevation in boiling point `K_(b)=` Molal elevation constant, `m=` Molality
Substitution `&` Calculation
`DeltaT_(b)=0.512xx0.322=0.165K`
Boiling point of solution is `(373+0.165)=373.165K`
`ul("Calculation of lowering of vapour pressure")`
Formulae: `(P_(A)^(0)-P_(A))/(P_(A)^(0))=x_(B)`
Explanation: `P_(A)^(0)=` Vapour pressure of pure solvent
`P_(A)=` Vapour pressure of soltuion
`X_(B)=` Mole friction of solute
Calculation of `X_(B)`
`X_(B)=(m)/(m+(1000)/(18))=(0.322)/(55.87)`
Substitution in formulae
`1-(P_(A))/(P_(A)^(0))=(0.322)/(55.87) rArr (P_(A))/(23.757)=1-(0.322)/(55.87)=(55.55)/(55.87)`
`P_(A)=(23.757xx55.55)/(55.87)=23.62mm of Hg`.