How many grams of solute (molecular weight $60$) are required to be dissolved in $180 \ g$ of water to reduce the vapour pressure to $\frac{5}{6}$ of the pure water?

$A$ solution is made by mixing $1 \ mole$ of volatile liquid $A$ with $3 \ moles$ of volatile liquid $B$. The vapour pressure of pure $A$ is $200 \ mm \ Hg$ and that of the solution is $500 \ mm \ Hg$. The vapour pressure of pure $B$ and the least volatile component of the solution,respectively,are $:$

Relative lowering of vapour pressure of a dilute solution is $0.5$. What is the mole fraction of the non-volatile solute?

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

Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.

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: