Question

Cyclohexane and ethanol in a particular mixture have vapour pressure of 376 mm when mole fraction of cyclohexane is 0.32. If vapour pressure of cyclohexane and ethanol are 280 mm and 168 mm respectively at the same temperature, mixture will be an ideal solution.

If true enter 1 else 0.

Answer 1

Raoult's Law states that the partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

In a mixture of liquids of A and B,

$p_A=x_A\times {p_A}^0$

$p_B=x_B\times {p_B}^0$

Here, $p_A$ and $p_B$ are the partial vapour pressures of the components A and B. In any mixture of gases, each gas exerts its own pressure. This is called its partial pressure and is independent of the other gases present. Even if you take all the other gases away, the remaining gas would still be exerting its own partial pressure.

The total vapour pressure of the mixture is equal to the sum of the individual partial pressures.

$p=p_A+p_{B}p=x_A.{p_A}^0+x_B.{p_B}^0$

Given,

${p_A}^0=280$ mm of Hg

${p_B}^0=168$ mm of Hg

Let mole fraction of A be $x_A$ and mole fraction of B will be $1-x_A$.

Then,

$p=280\times 0.32+168(1-0.32)$

$p=203.84mm$

The resultant pressure is $203.84mm$ when we are applying Raoult's law but the given pressure of the solution is different from the pressure of the Raoult's law.

Hence, the above solution is not obeying Raoult's law.