The vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?

In a solution,the mole fraction of solute is $0.2$ when the lowering in vapour pressure is $10 \text{ mm Hg}$. To achieve a lowering of vapour pressure of $20 \text{ mm Hg}$,the mole fraction of the solute in the solution should be:

Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $298\,K$ are $200\,mm\,Hg$ and $415\,mm\,Hg$ respectively. $(i)$ Calculate the vapour pressure of the solution prepared by mixing $25.5\,g$ of $CHCl_{3}$ and $40\,g$ of $CH_{2}Cl_{2}$ at $298\,K$ and,$(ii)$ mole fractions of each component in vapour phase.

The vapour pressure of benzene at $90\,^oC$ is $1020 \, \text{torr}$. $5 \, g$ of a solute is dissolved in $58.5 \, g$ of benzene. The vapour pressure of the solution is $990 \, \text{torr}$. What is the molar mass of the solute?

The vapour pressures of two volatile liquids $A$ and $B$ at $25^{\circ} C$ are $50 \ Torr$ and $100 \ Torr,$ respectively. If the liquid mixture contains $0.3$ mole fraction of $A$,then the mole fraction of liquid $B$ in the vapour phase is $\frac{x}{17}$. The value of $x$ 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 $Hg$?