What is the vapour pressure of a solution containing $0.1 \text{ mol}$ of non-volatile solute dissolved in $16.2 \text{ g}$ water (in $\text{ mmHg}$)? ($P_1^0 = 24 \text{ mmHg}$, molar mass of water $18 \text{ g mol}^{-1}$)

$160 \ g$ 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 '$A$' = $160 \ g \ mol^{-1}$)

In a mixture of liquids $A$ and $B$, if the mole fractions of component $A$ in vapour phase and liquid mixture are $x_1$ and $x_2$ respectively, then the total vapour pressure of liquid mixture is (where $P_{A}^{\circ}$ and $P_{B}^{\circ}$ are the vapour pressures of pure $A$ and $B$)

The vapour pressure of a solvent decreases by $2.5 \ mm \ Hg$ by adding a solute. What is the mole fraction of solute? (Vapour pressure of pure solvent is $250 \ mm \ Hg$)

Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)