How many grams of sucrose must be added to `360g` of water to lower the vapour pressure by `1.19mm Hg` at a temperatue at which vapour pressure of pure water is `25 mmHg` ?

Answer

`342 g.

Explanation

Given `DeltaP=1.19mm Hg,P_("solvent")=25mm Hg`
By Raoult's law
`(DeltaP)/(P_("solvent")=X_("solute")`
`tehrefore X_("solvent")=1-1.0476=0.9524`
`(n_("solvent")/(n_("total")=0.9524`
`n_("solvent")=(360)/(18)=20`
`therefore n_("total")=(n_("solvent"))/(0.9524)=(20)/(0.9524)=21`
`therefore n_("solute")=21-20=1xx342g=342g`.