$2 \ moles$ 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: $P_A^o = 120 \ torr, P_B^o = 80 \ torr$

Which of the following form an ideal solution?
$I$. Chloroethane and bromoethane
$II$. Benzene and toluene
$III$. $n$-hexane and $n$-heptane
$IV$. Phenol and aniline

At $373 \ K$,heptane and octane form an ideal solution. The vapor pressures of the two pure liquid components (heptane and octane) are $105 \ kPa$ and $45 \ kPa$,respectively. The vapor pressure of the solution formed by mixing $25 \ g$ of heptane and $35 \ g$ of octane is ........ $kPa$. (Molar masses of heptane and octane are $100 \ g \ mol^{-1}$ and $114 \ g \ mol^{-1}$,respectively.)

$A$ liquid mixture is an ideal solution,if
$(I)$ It obeys ideal gas equation
$(II)$ It obeys Raoult's law at all concentrations
$(III)$ Solute - solute,solute - solvent and solvent - solvent interactions are similar

At $300 \ K$,the vapour pressures of $A$ and $B$ liquids are $500 \ mm \ Hg$ and $400 \ mm \ Hg$ respectively. Equal moles of $A$ and $B$ are mixed to form an ideal solution. The mole fraction of $A$ and $B$ in the vapor state is respectively:

What is a non-ideal solution and what is meant by positive deviation from Raoult's law?