(a) Why is the vapour pressure of a solution of glucose in water lower than that of water ?
(b) A `6.90M` solution of `KOH` in water contains `30%` by mass of `KOH`. Calculate the density of the `KOH` solution. [Molar mass of `KOH=56g "mol"^(-1)]`

Answer

(a) When glucose is dissolved in water, the surface molecules consist of both glucose and water, therefore, escaping tendency of molecules of solvent in vapour decreases, therefore, vapour pressure decreases.
(b) `M=(% "by massxxdxx10)/("Mol wt.") rArr 6.90=(30xxdxx10)/(56)`
`rArr d=(6.90xx56)/(30xx10)=(386.4)/(300)=1.288g cm^(-3)`