The vapour pressure of pure benzene and toluene are $0.9 \ Bar$ and $0.85 \ Bar$ respectively. Calculate the vapour pressure of the solution prepared by dissolving $7.8 \ g$ of benzene in $180 \ g$ of toluene. (in $Bar$)

The plot of total vapour pressure as a function of mole fraction of the components of an ideal solution formed by mixing liquids $X$ and $Y$ is

In a solution,the mole ratio of pentane $(A)$ and hexane $(B)$ is $1:4$. At $20^o C$,the vapor pressures of these pure hydrocarbons are $440 \ mm$ and $120 \ mm$ respectively. The mole fraction of pentane in the vapor phase is.....

At $298 \ K$,vapour pressures of two pure liquids $A$ and $B$ are $200 \ mm \ Hg$ and $400 \ mm \ Hg$ respectively. If mole fractions of $A$ and $B$ in solution are $0.7$ and $0.3$ respectively,what is the mole fraction of $B$ in vapour phase?

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. If the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$,then the mole fraction of $B$ in the solution at the same temperature is

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