$V.P.$ of pure $A$: $p^o_A = 100 \, mmHg$
$V.P.$ of pure $B$: $p^o_B = 150 \, mmHg$
$A$ solution containing $2 \, moles$ of $A$ and $3 \, moles$ of $B$ will have a total vapour pressure of approximately $......... \, mmHg$.

The number of moles of chlorobenzene and bromobenzene are $0.1$ and $0.2$ respectively. The vapour pressure of pure chlorobenzene and pure bromobenzene are $0.350 \ bar$ and $0.500 \ bar$ respectively. Calculate the total vapour pressure of the solution prepared by mixing these two components.

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)

At a certain temperature,the mole fraction of a solute is $0.15$ and the vapor pressure of the pure solvent is $120 \, torr$. If the solute is non-volatile,what will be the relative lowering in vapor pressure?

Vapour pressure of chloroform $(CHCl_3)$ and dichloromethane $(CH_2Cl_2)$ at $25^\circ C$ are $200 \ mm \ Hg$ and $41.5 \ mm \ Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \ g$ of $CHCl_3$ and $40 \ g$ of $CH_2Cl_2$ at the same temperature will be (Molecular mass of $CHCl_3 = 119.5 \ u$ and molecular mass of $CH_2Cl_2 = 85 \ u$)

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 $Hg$?