"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

$A$ gaseous mixture of two substances $A$ and $B$,under a total pressure of $0.8 \ atm$ is in equilibrium with an ideal liquid solution. The mole fraction of substance $A$ is $0.5$ in the vapour phase and $0.2$ in the liquid phase. The vapour pressure of pure liquid $A$ is $...... \ atm$. (Nearest integer)

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?

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

The vapor pressures of two liquids $A$ and $B$ are $80 \ mm$ and $60 \ mm$ respectively. If $3 \ mol$ of $A$ and $2 \ mol$ of $B$ are mixed,the total vapor pressure of the resulting solution will be .......... $mm$.

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. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?