At $T \ K$,the vapour pressure of an $x$ molal aqueous solution containing a non-volatile solute is $12.078 \ kPa$. The vapour pressure of pure water at $T \ K$ is $12.3 \ kPa$. What is the value of $x$?

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

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 $(Y_{A})$?

What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)

The mass of non-volatile,non-electrolyte solute (molar mass $= 50 \ g \ mol^{-1}$) needed to be dissolved in $114 \ g$ octane to reduce its vapour pressure to $75\%$ is .............. $g$.

$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)