Calculate the relative lowering of vapour pressure of a solution containing $46 \ g$ of non-volatile solute in $162 \ g$ of water at $20^{\circ} C$. [Molar mass of non-volatile solute $= 46 \ g \ mol^{-1}$]

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

According to Raoult's law,the relative lowering of vapour pressure for a solution is equal to which of the following?

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 $CCl_4 = 1.58\,g/cm^3$)