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

At $363 \ K,$ the vapour pressure of $A$ is $21 \ kPa$ and that of $B$ is $18 \ kPa$. One mole of $A$ and $2 \ moles$ of $B$ are mixed. Assuming that this solution is ideal,the vapour pressure of the mixture is $....... \ kPa$. (Round off to the Nearest Integer).

Liquids $A$ and $B$ form an ideal solution. At $30\,^oC$,the total vapour pressure of a solution containing $1\,mol$ of $A$ and $2\,mol$ of $B$ is $250\,mm\,Hg$. The total vapour pressure becomes $300\,mm\,Hg$ when $1$ more $mol$ of $A$ is added to the first solution. The vapour pressures of pure $A$ and $B$ at the same temperature are

$A$ liquid solution is formed by mixing $10 \, moles$ of aniline and $20 \, moles$ of phenol at a temperature where the vapour pressure of pure liquid aniline and phenol are $90 \, mmHg$ and $87 \, mmHg$ respectively. The possible vapour pressure of the solution at that temperature is ............. $mmHg$.

Identify the mixture that shows positive deviations from Raoult's Law.

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):