$A$ 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)

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 $g$)

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

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 vapor pressures of two liquids $P$ and $Q$ are $80 \ torr$ and $60 \ torr$,respectively. The total vapor pressure of a solution formed by mixing $3 \ moles$ of $P$ and $2 \ moles$ of $Q$ is .......... $torr$.

The vapor pressure of water at $23\,^{\circ}C$ is $19.8\,mm\,Hg$. What will be the vapor pressure (in $mm\,Hg$) of the resulting solution when $0.1\,mol$ of glucose is dissolved in $178.2\,g$ of water?