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 $(Y_{A})$?

At a temperature of $300 \ K$,if a liquid is transferred from a small test tube to a large beaker,what will be the change in its vapor pressure?

Two liquids $X$ and $Y$ form an ideal solution. At a constant temperature of $300 \ K$,the vapor pressure of a solution containing $1 \ mol$ of $X$ and $3 \ mol$ of $Y$ is $550 \ mm \ Hg$. If $1 \ mol$ of $Y$ is added to this solution,the vapor pressure of the solution increases by $10 \ mm \ Hg$. What are the vapor pressures of $X$ and $Y$ in their pure states?

If $p^{\circ}$ and $p$ are the vapour pressure of the pure solvent and solution, and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution, then the correct relation between $p$ and $p^{\circ}$ is:

The relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute. This statement was given by whom?

What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water $\left(P_1^0 = 32 \ mm \ Hg\right) (in $mm \ Hg$)?$