The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600\, mm\, Hg$,respectively at $298\, K$. On mixing the two liquids,the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution,the mole fractions of components $A$ and $B$ in vapour phase,respectively are

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

The variation of vapour pressure of different liquids with temperature is shown in the figure.
$(i)$ Calculate graphically the boiling points of liquids $A$ and $B$.
(ii) If we take liquid $C$ in a closed vessel and heat it continuously,at what temperature will it boil?
(iii) At high altitude,atmospheric pressure is low (say $60 \ mm \ Hg$). At what temperature does liquid $D$ boil?
(iv) $A$ pressure cooker is used for cooking food at hill stations. Explain in terms of vapour pressure why this is so.

Calculate the mole fraction of solute in a solution if the vapour pressure of the solution and the pure solvent are $27 \text{ mm Hg}$ and $30 \text{ mm Hg}$ respectively at $298 \text{ K}$.

$18 \ g$ 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: