Mixing $1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$ forms an ideal solution. The vapor pressure of the solution will be .......... $mm \ Hg$. (in $.2$)

An example of a near-ideal solution is:

For a solution of two liquids $A$ and $B$,if $P = X_A (P_A^o - P_B^o) + P_B^o$ holds true,then what type of solution is it?

The total pressure observed by mixing two liquids $A$ and $B$ is $350 \ mm \ Hg$ when their mole fractions are $0.7$ and $0.3$ respectively. The total pressure becomes $410 \ mm \ Hg$ if the mole fractions are changed to $0.2$ and $0.8$ respectively for $A$ and $B$. The vapour pressure of pure $A$ is $........... \ mm \ Hg$. (Nearest integer)
Consider the liquids and solutions behave ideally.

$A$ solution of two miscible liquids showing negative deviation from Raoult's law will have :

On mixing,heptane and octane form an ideal solution. At $373 \ K$,the vapour pressures of the two liquid components (heptane and octane) are $105 \ kPa$ and $45 \ kPa$ respectively. Vapour pressure of the solution obtained by mixing $25.0 \ g$ of heptane and $35 \ g$ of octane will be $........ \ kPa$.
(molar mass of heptane $= 100 \ g \ mol^{-1}$ and of octane $= 114 \ g \ mol^{-1}$)