Mixing $1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$ forms an ideal solution. The vapor pressure of the solution will be .......... $mm \ Hg$. (in $.2$)

Assuming the formation of an ideal solution,determine the boiling point of a mixture containing $1560 \ g$ benzene (molar mass $= 78 \ g/mol$) and $1125 \ g$ chlorobenzene (molar mass $= 112.5 \ g/mol$) against an external pressure of $1000 \ torr$ using the provided graph. (in $^{\circ}C$)

Explain non-ideal solutions and the occurrence of positive and negative deviations in non-ideal solutions.

Two liquids $A$ and $B$ form an ideal solution. When they are mixed in a molar ratio of $1:1$,the vapor pressure of the solution at $300 \ K$ is $400 \ mm \ Hg$. When they are mixed in a molar ratio of $1:2$,the vapor pressure of the solution at the same temperature is $350 \ mm \ Hg$. The vapor pressures of pure liquids $A$ and $B$ are respectively:

$2 \ moles$ each of liquids $A$ and $B$ are dissolved to form an ideal solution. What will be the mole fraction of $B$ in the vapour phase? Given: $P_A^o = 120 \ torr, P_B^o = 80 \ torr$

$A$ solution of two miscible liquids showing negative deviation from Raoult's law will have :