At $350 \ K$,the vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively. If the total vapour pressure of the liquid mixture is $600 \ mm \ Hg$,the mole fractions of $A$ and $B$ in the vapour phase respectively are:

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)

The vapour pressure of pure benzene at a certain temperature is $0.850 \ bar$. $A$ non-volatile,non-electrolyte solid weighing $0.5 \ g$ is added to $39.0 \ g$ of benzene (molar mass $78 \ g/mol$). The vapour pressure of the solution then is $0.845 \ bar$. What is the molecular mass of the solid substance?

Calculate the molar mass of a nonvolatile solute if $6.4 \text{ g}$ of it dissolved in $100 \text{ g}$ water produces a relative lowering in vapour pressure of $0.016$ at $300 \text{ K}$.

Explain how the solubility of a gaseous solute and the vapour pressure of a liquid solution follow Henry's and Raoult's laws.

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.