What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

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

The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600\, mm\, Hg$,respectively at $298\, K$. On mixing the two liquids,the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution,the mole fractions of components $A$ and $B$ in vapour phase,respectively are

An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004 \, bar$ at the normal boiling point of the solvent. The molar mass of the solute is $......... \, g/mol$.

At $300 \ K$, the vapour pressure of toluene and benzene are $3.63 \ kPa$ and $9.7 \ kPa$ respectively. What is the composition of vapour in equilibrium with the solution containing $0.4$ mole fraction of toluene? (Assume the solution is ideal)