When $0.25 \text{ moles}$ of a non-volatile, non-ionizable solute was dissolved in $1 \text{ mole}$ of a solvent, the vapor pressure of the solution was $x\%$ of the vapor pressure of the pure solvent. What is $x$ (in $\%$)?

According to Raoult's law,the relative lowering of vapour pressure of a solution of a non-volatile solute is equal to:

At $T \ K$,the vapour pressure of water is $x \ kPa$. What is the vapour pressure (in $kPa$) of a $1 \ m$ (molal) aqueous solution containing a non-volatile solute (in $x$)?

Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.
Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.

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 $Hg$?

$18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water. The vapour pressure of water for this aqueous solution at $100^\circ C$ is ........ $Torr$.