In a liquid mixture of $A$ and $B$,what is the mole fraction of $A$ in the vapor phase $(X_A = 0.4)$? Given: $P_A^o = 100 \, mm \, Hg$ and $P_B^o = 200 \, mm \, Hg$.

$18 \ g$ glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ water. The vapour pressure of this aqueous solution at $100 \ ^{\circ}C$ (in $torr$) is:

The vapour pressure of pure benzene and methyl benzene at $27^{\circ} C$ is given as $80 \ Torr$ and $24 \ Torr$,respectively. The mole fraction of methyl benzene in vapour phase,in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is. . . . . . .$\times 10^{-2}$ (nearest integer)

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

$x \ g$ of a solute is dissolved in $y \ g$ of two different liquids $A$ and $B$. The relative lowering of vapor pressure of the solution in $A$ is twice that of the solution in $B$. If the moles of solute are negligible compared to the moles of solvent,which of the following statements regarding the molar masses $M_A$ and $M_B$ of the solvents is correct?

If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?