Liquids $A$ and $B$ form an ideal solution. The vapour pressure of pure $A$ and $B$ are $50 \ mm \ Hg$ and $32 \ mm \ Hg$ respectively at $300 \ K$. One mole of liquid $A$ is mixed with $1$ mole of liquid $B$. What is the approximate mole fraction of $A$ in the vapour phase?

$A$ solution of acetone in ethanol:

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:

$A$ mixture of toluene and benzene boils at $100^{\circ} C$. Assuming ideal behaviour,the mole fraction of toluene in the mixture is closest to (Vapour pressures of pure toluene and pure benzene at $100^{\circ} C$ are $0.742 \ bar$ and $1.800 \ bar$,respectively. $1 \ atm = 1.013 \ bar$)

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?

Which statement about the composition of vapour over an ideal $1:1$ molar mixture of benzene and toluene is correct? Assume the temperature is constant at $25\,^{\circ}C$. Vapour pressure data $(25\,^{\circ}C)$:
Benzene: $75\, mm\, Hg$
Toluene: $22\, mm\, Hg$