$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.

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 ........ $torr$.

At $80\,^oC,$ the vapour pressure of pure liquid $'A'$ is $520\, mm\,Hg$ and that of pure liquid $'B'$ is $1000\, mm\,Hg.$ If a mixture solution of $'A'$ and $'B'$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $'A'$ in the mixture is ........... $mol$ percent $(1\, atm = 760\, mm\,Hg).$

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

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 $Bar$)