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 $\%$)?

When a solute is added to a solvent having a vapor pressure of $0.80 \ atm$,the vapor pressure decreases to $0.60 \ atm$. The mole fraction of the solute is .........

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 150 - 60 X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and pure $B$ at the same temperature are respectively:

At $25^\circ C$,the vapor 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 $CCl_4$,the vapor pressure of the resulting solution is $...... \ mm \ Hg$. (Density of $CCl_4 = 1.58 \ g/cm^3$)

At a certain temperature,the mole fraction of a solute is $0.15$ and the vapor pressure of the pure solvent is $120 \, torr$. If the solute is non-volatile,what will be the relative lowering in vapor pressure?