Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

Calculate the mass of a non-volatile solute (molar mass $40 \, g \, mol^{-1}$) which should be dissolved in $114 \, g$ octane to reduce its vapour pressure to $80 \%$.

Derive an equation for a solution that shows the relation between total pressure and the mole fraction of a volatile solute and a volatile solvent,and explain it by plotting a graph.

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

At $25\, ^oC$,benzene and toluene form an ideal solution. What will be the vapor pressure (in $mm \, of \, Hg$) of a solution containing benzene and toluene in a weight ratio of $1:2$? The vapor pressures of pure benzene and pure toluene are $75\, mm \, of \, Hg$ and $30\, mm \, of \, Hg$,respectively.

The vapour pressure of a dilute aqueous solution of glucose is $750 \ mm$ of mercury at $373 \ K$. The mole fraction of solvent is