The vapour pressure of pure liquid $\mathrm{A}$ is $10 \mathrm{torr}$ and at the same temperature when $1 \mathrm{g}$ of $\mathrm{B}$ solid is dissolved in $20 \mathrm{g}$ of $\mathrm{A}$, its vapour pressure is reduced to $9.0 \mathrm{torr}$. If the molecular mass of $\mathrm{A}$ is $200 \mathrm{amu}$, then the molecular mass of $\mathrm{B}$ is:

A mixture of toluene and benzene boils at $100°\mathrm{C}$. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to:

[Vapour pressures of pure toluene and pure benzene at $100°\mathrm{C}$ are $0.742$ and $1.800 \mathrm{bar}$, respectively. $1 \mathrm{atm}=1.013 \mathrm{bar}$]

Two liquids, $\mathrm{X}$ and $\mathrm{Y}$, form an ideal solution. At $300 \mathrm{K}$, the vapour pressure of the solution containing one mole of $\mathrm{X}$ and three moles of $\mathrm{Y}$ is $550 \mathrm{mm} \mathrm{Hg}$ . At the same temperature, if one mole of $\mathrm{Y}$ is further added to this solution, the vapour pressure of the solution increases by $10 \mathrm{mm} \mathrm{Hg}$. The vapour pressure of $\mathrm{X}$ and $\mathrm{Y}$ (in $\mathrm{mm} \mathrm{Hg}$) in their pure states will be, respectively:

Vapour pressure of a highly dilute solution of a sparingly soluble salt $\mathrm{AB}$ is $20 \mathrm{mm}$ of $\mathrm{Hg}$ at temperature $\mathrm{T},$ while pure water exerts a pressure of $20.005 \mathrm{mm}$ $\mathrm{Hg}$ at the same temperature. The solubility product constant of $\mathrm{AB}$ at the same temperature is approximately: