At a certain temperature,when some amount of glucose is dissolved in water,it is observed that the lowering in vapour pressure is $0.6 \ mm \ Hg$ for this solution. What will be the vapour pressure of the same glucose solution if its molality is $(\frac{1}{18}) \ mol \ kg^{-1}$ (in $mm \ Hg$)?

The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

Vapour pressure of a pure solvent at $90\,^{\circ}C$ is $1020\, torr$. $A$ solution of $0.1\, m$ concentration has a vapour pressure of $1000\, torr$. What is the molecular weight of the solvent?

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

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).