The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

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 $\%$)?

If $x \ g$ of a non-volatile,non-electrolytic solute is dissolved in $114 \ g$ of octane to reduce its vapour pressure to $80 \%$ of that of pure octane,then $x$ is ........... $g$. (Molar mass of solute $= 40 \ g/mol$)

The vapour pressure of ethanol $(C_2H_5OH)$ and methanol $(CH_3OH)$ at $298 \ K$ is $44.5 \ mm \ Hg$ and $88.7 \ mm \ Hg$ respectively. At the same temperature,an ideal solution is prepared by mixing $60 \ g$ of ethanol and $40 \ g$ of methanol. Find the mole fraction of methanol in the vapour state.

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 $mm \ Hg$)

The relative lowering of the vapour pressure is equal to the ratio between the number of