The vapour pressure of an aqueous solution of methanol is:

The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

What is the vapour pressure of a solution when $2 \ mol$ of a non-volatile solute are dissolved in $20 \ mol$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

If $P^o$ and $P$ are the vapour pressure of a solvent and its solution respectively and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then the correct relation is:

An aqueous solution of glucose was prepared by dissolving $18 \ g$ of glucose in $90 \ g$ of water. The relative lowering in vapour pressure is