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 $18 \text{ g mol}^{-1}$)

The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

Vapour pressure of a pure solvent is $550 \ mm$ of $Hg$. By addition of a non-volatile solute,it decreases to $510 \ mm$ of $Hg$. Calculate the mole fraction of the solute in the solution.

What is the mole ratio of benzene $(P_B^o = 150\, torr)$ and toluene $(P_T^o = 50\, torr)$ in the vapour phase if the given solution has a vapour pressure of $120\, torr$?

$A$ solution was prepared by dissolving $0.1 \ mol$ of a nonvolatile solute in $0.9 \ mol$ of water. What is the relative lowering of vapour pressure of the solution?