Which solution has the lowest vapor pressure?

$A$ gaseous mixture of two substances $A$ and $B$,under a total pressure of $0.8 \ atm$ is in equilibrium with an ideal liquid solution. The mole fraction of substance $A$ is $0.5$ in the vapour phase and $0.2$ in the liquid phase. The vapour pressure of pure liquid $A$ is $...... \ atm$. (Nearest integer)

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)

The ratio of loss in weight of solvent to gain in weight of $CaCl_{2}$ tube in Ostwald-Walker method is:

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)