$3 \ g$ of urea is dissolved in $45 \ g$ of water. The relative lowering of vapour pressure 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 $(Y_{A})$?

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?

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 pressures of $A$ and $B$ at $25^{\circ} C$ are $90 \ mm \ Hg$ and $15 \ mm \ Hg$ respectively. If $A$ and $B$ are mixed such that the mole fraction of $A$ in the mixture is $0.6$,then the mole fraction of $B$ in the vapour phase is $x \times 10^{-1}$. The value of $x$ is $.....$ (Nearest integer)

The vapour pressure of pure water at $25\,^{\circ}C$ is $23.62\, mm$. What will be the vapour pressure of a solution of $1.5\, g$ urea in $50\, g$ water?