At $88\,^oC$,the vapor pressure of benzene is $900 \, \text{torr}$ and that of toluene is $360 \, \text{torr}$. What is the mole fraction of benzene in the mixture with toluene that will boil at $88\,^oC$ and $1 \, \text{atm}$ pressure,assuming it forms an ideal solution?

If vapour pressures of pure liquids $A$ and $B$ are $300$ and $800 \, torr$ respectively at $25 \, ^{\circ}C$. When these two liquids are mixed at this temperature to form a solution in which mole percentage of $B$ is $92$,then the total vapour pressure is observed to be $0.95 \, atm$. Which of the following is true for this solution?

Which of the following is not correct?

Liquid $A$ and $B$ form an ideal solution. The vapour pressure of pure liquids $A$ and $B$ are $350 \ mm \ Hg$ and $750 \ mm \ Hg$ respectively at the same temperature. If $x_A$ and $x_B$ are the mole fractions of $A$ and $B$ in the solution,and $y_A$ and $y_B$ are the mole fractions of $A$ and $B$ in the vapour phase,then:

Two statements are given below:
Statement-$I$: Liquids $A$ and $B$ form a non-ideal solution with positive deviation. The interactions between $A$ and $B$ are weaker than $A-A$ and $B-B$ interactions.
Statement-$II$: For an ideal solution,$\Delta_{mix} H = 0$ and $\Delta_{mix} V = 0$.
The correct answer is:

For which of the following liquid mixtures $\Delta_{\text{mix}}H=0$ and $\Delta_{\text{mix}}V=0$?