How many grams of `NaBr` must be added to `270g` of water to lower the vapour pressure by `3.125mm Hg` at which vapour pressure of water is `50mm Hg (Na=23,Br=80)` ? Assume `100%` ionisation of `NaBr`.

Answer

`51.5g`.

Explanation

`NaBrhArrNa^(+)+Br^(-)`
van'tHoff factor `i=2`
let moles of `NaBr=2n_(1)`
mole of `H_(2)O(mn_(2))=(270)/(18)=15`
`(DeltaP)/(P_("solvent")=X_("solute")=(n_(1)i)/(n_(1)i+n_(2))`
`(3.125)/(50)=(2n_(1))/(2n_(1)+15)`
`(2n_(1)+15)/(2n_(1))=(50)/(3.125)=16`
`1+(15)/(2n_(1))=16`
`(15)/(2n_(1))=15`
`n_(1)=0.5`
`therefore NaBr` taken `=0.5` mol `=0.5xx103=51.5g`.