The correct option for the value of vapour pressure of a solution at $45^{\circ} C$ with benzene to octane in molar ratio $3 : 2$ is ...... $mm$ of $Hg$. [At $45^{\circ} C$,vapour pressure of benzene is $280 \ mm \ Hg$ and that of octane is $420 \ mm \ Hg$. Assume ideal solution.]

Two liquids $A$ and $B$ form an ideal solution. At $320 \ K$, the vapour pressure of the solution, containing $3 \ mol$ of $A$ and $1 \ mol$ of $B$ is $500 \ mm \ Hg$. At the same temperature, if $1 \ mol$ of $A$ is further added to this solution, the vapour pressure of the solution increases by $20 \ mm \ Hg$. The vapour pressure (in $mm \ Hg$) of $B$ in the pure state is . . . . . . (Nearest integer).

The vapor pressure of a pure liquid solvent $(X)$ decreases from $0.80 \ atm$ to $0.60 \ atm$ upon the addition of a non-volatile solute $(Y)$. What is the mole fraction of $(Y)$ in the solution?

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?

The vapour pressure of $30 \%$ $(w/v)$ aqueous solution of glucose is $...... \ mm \ Hg$ at $25^{\circ} \ C$. [Given : The density of $30 \%$ $(w/v)$ aqueous solution of glucose is $1.2 \ g \ cm^{-3}$ and vapour pressure of pure water is $24 \ mm \ Hg$.] (Molar mass of glucose is $180 \ g \ mol^{-1}$.)

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 $Hg$?