At a particular temperature,the vapour pressures of two liquids $A$ and $B$ are respectively $120 \, mm$ and $180 \, mm$ of mercury. If $2 \, moles$ of $A$ and $3 \, moles$ of $B$ are mixed to form an ideal solution,the vapour pressure of the solution at the same temperature will be (in $mm$ of mercury):

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.)

Liquids $A$ and $B$ form an ideal solution. The vapour pressure of pure $A$ and $B$ are $50 \ mm \ Hg$ and $32 \ mm \ Hg$ respectively at $300 \ K$. One mole of liquid $A$ is mixed with $1$ mole of liquid $B$. What is the approximate mole fraction of $A$ in the vapour phase?

Explain the terms ideal and non-ideal solutions in the light of forces of interactions operating between molecules in liquid solutions.

$A$ mixture of phenol and aniline shows negative deviation from Raoult's law. This is due to the formation of

In a mixture of $A$ and $B$ components,the solution shows positive deviation from Raoult's law when: