If the total vapour pressure of the liquid mixture $A$ and $B$ is given by the equation: $P = 180X_A + 90 \, mm \, Hg$,then the ratio of the vapour pressure of the pure liquids $A$ and $B$ is given by:

Dry air was drawn through bulbs containing pure water,then through bulbs containing a solution of $600 \ g$ of a non-electrolyte in $360 \ g$ of water at the same temperature,and finally through a tube in which dried $CaCl_2$ was placed. The solution bulb gained $1.5 \ g$ and the dried $CaCl_2$ gained $2 \ g$. The molecular mass of the solute is:

Dry air was passed successively through a solution of $5 \ g$ of a solute in $80 \ g$ of water and then through pure water. The loss in weight of solution was $2.50 \ g$ and that of pure solvent $0.04 \ g$. What is the molecular weight of the solute?

Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)

If vapour pressure of pure solvent and solution are $240 \ mm Hg$ and $216 \ mm Hg$ respectively,then the mole fraction of solvent in the solution is:

Explain how the solubility of a gaseous solute and the vapour pressure of a liquid solution follow Henry's and Raoult's laws.