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

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 $Hg$?

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 $25\,^oC$,the vapour pressure of $CCl_4$ is $143\,mm\,Hg$. If $0.5\,g$ of a non-volatile solute (molar mass $= 65\,g/mol$) is dissolved in $100\,mL$ of $CCl_4$,what will be the vapour pressure of the solution? (Density of $CCl_4 = 1.58\,g/cm^3$)

Two beakers $A$ and $B$ containing pure water and an aqueous solution,respectively,are placed in a closed container. What will be observed gradually with the passage of time?

The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution,which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)