For a dilute solution containing 2.5 g of a non-voltile and non-electrolyte solute in 100 g of water, the elevation in boiling at 1 atm pressure is `2^(@)C`. Assuming that the concentration of solute is mich less lower that the concentration of the solvent, what is the capour pressure (mm of Hg) of the solution? Given `K_(b)=0.76 K kg mol^(-1)`)

Explanation

Calcultion of the molar mass of solute.
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))`
`K_(b)=0.76 K kg mol^(-1),,W_(B)=2.5g, W_(A)=100 g= 0.1 kg, DeltaT_(b)=2K`
`M_(B)=((0.76K mol^(-1)xx(2.5g)))/((2K)xx(0.1 Kg))=9.5 g mol^(-1)`
Calculationof the vapour pressure of solution
`(P_(A)^(0)-P_(S))/P_(S)=x_(B)=(W_(B)/M_(B))/(W_(B)/M_(B)+W_(A)/M_(A))=((2.5g)/((0.5g mol^(-1))))/((2.5g)/((9.5. g mol^(-1)))+(100g)/((18g mol^-1)))=0.051`
`((760mm)-P_(s))/P_(s)=0.051 or ((760))/P_(s)-1=0.051`
` (670mm)/P_(s)=1.051=1.051 or P_(s)=((760mm))/((1.051))=723.12mm`
`"Vapour pressure of solution"=723.12`