"Relative lowering in vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute". The above statement is:

The vapour pressure of acetone at $20\,^{\circ}C$ is $185\,torr.$ When $1.2\,g$ of a non-volatile substance was dissolved in $100\,g$ of acetone at $20\,^{\circ}C,$ its vapour pressure was $183\,torr.$ The molar mass $(g\,mol^{-1})$ of the substance is:

Two liquids $A$ and $B$ form an ideal solution. At $320 \ K$,the vapour pressure of the solution,containing $3 \ mol$ of $A$ and $1 \ mol$ of $B$ is $500 \ mm \ Hg$. At the same temperature,if $1 \ mol$ of $A$ is further added to this solution,the vapour pressure of the solution increases by $20 \ mm \ Hg$. The vapour pressure (in $mm \ Hg$) of $B$ in the pure state is . . . . . . (Nearest integer).

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 $mm \ Hg$)

In a solution,the mole ratio of pentane $(A)$ and hexane $(B)$ is $1:4$. At $20^o C$,the vapor pressures of these pure hydrocarbons are $440 \ mm$ and $120 \ mm$ respectively. The mole fraction of pentane in the vapor phase is.....

What is the mole ratio of benzene $(P_B^o = 150\, torr)$ and toluene $(P_T^o = 50\, torr)$ in the vapour phase if the given solution has a vapour pressure of $120\, torr$?