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 $m$)

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

$3 \ g$ of urea is dissolved in $45 \ g$ of water. The relative lowering of vapour pressure is

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm$ and $700 \ mm$ of $Hg$ at $350 \ K$ respectively. If the total vapour pressure of the mixture is $600 \ mm$ of $Hg$,the composition of the mixture in the solution is

The boiling points of $C_6H_6$,$CH_3OH$,$C_6H_5NH_2$,and $C_6H_5NO_2$ are $80 \, ^\circ C$,$65 \, ^\circ C$,$184 \, ^\circ C$,and $212 \, ^\circ C$ respectively. Which one exhibits the maximum vapor pressure at room temperature?