At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

The vapour pressure of pure water at $25\,^{\circ}C$ is $23.62\, mm$. What will be the vapour pressure of a solution of $1.5\, g$ urea in $50\, g$ water?

Calculate the vapour pressure of a solution containing a mixture of $2 \ moles$ of volatile liquid $A$ and $3 \ moles$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (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 $Hg$?

Choose the correct statement with respect to the vapour pressure of a liquid among the following:

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