$A$ mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \, mm$ at $300 \, K$. The vapour pressure of propyl alcohol is $200 \, mm$. If the mole fraction of ethyl alcohol is $0.6$,its vapour pressure (in $mm$) at the same temperature will be:

The vapor pressure of a pure liquid solvent $(X)$ decreases from $0.80 \ atm$ to $0.60 \ atm$ upon the addition of a non-volatile solute $(Y)$. What is the mole fraction of $(Y)$ in the solution?

The vapour pressure of benzene at a certain temperature is $640 \ mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \ g$ is added to $39.08 \ g$ of benzene. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?

The vapour pressure of benzene at $90\,^oC$ is $1020 \, \text{torr}$. $5 \, g$ of a solute is dissolved in $58.5 \, g$ of benzene. The vapour pressure of the solution is $990 \, \text{torr}$. What is the molar mass of the solute?

"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:

For two volatile liquids $A$ and $B$,if the vapour pressure ratio of $P_{A}^{0} : P_{B}^{0} = 1 : 2$ and the mole fraction ratio in the liquid phase is $X_{A} : X_{B} = 1 : 2$,then find the mole fraction of component $A$ in the vapour state $(Y_{A})$?