The vapour pressure of two pure liquids $(A)$ and $(B)$ are $100 \ torr$ and $80 \ torr$ respectively. The total pressure of the solution obtained by mixing $2 \ mole$ of $(A)$ and $3 \ mole$ of $(B)$ would be ........ $torr$.

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)

Which of the following can be measured by the Ostwald-Walker dynamic method?

Vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature. The mole fraction of toluene in the vapour phase,which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$,will be:

The vapour pressure of pure liquids $A$ and $B$ are $450$ $mm$ $Hg$ and $700$ $mm$ $Hg$ respectively,at $350$ $K$. Find out the composition of the liquid mixture if the total vapour pressure is $600$ $mm$ $Hg$. Also,find the composition of the vapour phase.

$V.P.$ 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 $......... \, mmHg$.