How many grams of a non-volatile solute (molar mass $40 \ g/mol$) must be added to $114 \ g$ of octane to reduce its vapour pressure by $20\%$ (in $g$)?

The vapor pressure of water at $23\,^{\circ}C$ is $19.8\,mm\,Hg$. What will be the vapor pressure (in $mm\,Hg$) of the resulting solution when $0.1\,mol$ of glucose is dissolved in $178.2\,g$ of water?

If the total vapour pressure of the liquid mixture $A$ and $B$ is given by the equation: $P = 180X_A + 90 \, mm \, Hg$,then the ratio of the vapour pressure of the pure liquids $A$ and $B$ is given by:

Two open beakers,one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute,are together sealed in a container. Over time:

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

The vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?