$V.P.$ of pure $A$: $p^o_A = 100 \, mmHg$
$V.P.$ of pure $B$: $p^o_B = 150 \, mmHg$
$A$ solution containing $2 \, moles$ of $A$ and $3 \, moles$ of $B$ will have a total vapour pressure of approximately $......... \, mmHg$.

The vapour pressure of pure benzene and methyl benzene at $27^{\circ} C$ is given as $80 \ Torr$ and $24 \ Torr$,respectively. The mole fraction of methyl benzene in vapour phase,in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is. . . . . . .$\times 10^{-2}$ (nearest integer)

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

$100 \, g$ of liquid $A$ (molar mass $140 \, g \, mol^{-1}$) was dissolved in $1000 \, g$ of liquid $B$ (molar mass $180 \, g \, mol^{-1}$). The vapour pressure of pure liquid $B$ was found to be $500 \, torr$. Calculate the vapour pressure of pure liquid $A$ and its vapour pressure in the solution if the total vapour pressure of the solution is $475 \, torr$.

$A$ graph of vapour pressure and temperature for three different liquids $X$,$Y$ and $Z$ is shown below.
The following inferences are made:
$(A)$ $X$ has higher intermolecular interactions compared to $Y$.
$(B)$ $X$ has lower intermolecular interactions compared to $Y$.
$(C)$ $Z$ has lower intermolecular interactions compared to $Y$.
The correct inference$(s)$ is/are:

The vapour pressures of $A$ and $B$ at $25^{\circ} C$ are $90 \ mm \ Hg$ and $15 \ mm \ Hg$ respectively. If $A$ and $B$ are mixed such that the mole fraction of $A$ in the mixture is $0.6$,then the mole fraction of $B$ in the vapour phase is $x \times 10^{-1}$. The value of $x$ is $.....$ (Nearest integer)