I know that Raoult's law holds true only for a non volatile solute in a volatile solvent mixture wherein the vapour pressure of the solution gets lowered due to the addition of solute.

  1. Now, first of all does it hold for a mixture of immiscible liquids ?

  2. Does it hold for a mixture of immiscible liquids undergoing steam distillation? I came across that it does but can someone explain how ? I am just not able to understand.

My understanding :

(1) Raoult's law is used to calculate relative lowering of vapour pressure of solvent when a solute is added. In a mixture of immiscible liquids the vapour pressure of the solution increases and is equal to the vapour pressures of the two solvents independently. So there is no lowering in the vapour pressure of the solution. Based on this can I say that Raoult's law does not hold ?

(2) A mixture of immiscible liquids is not ideal right ? So Raoult's law does not hold.

(3) If the liquids are volatile, it won't hold.

But despite all this a problem was solved this way:

When a liquid that is immiscible with water was steam distilled at $\pu{95.2°C}$ at a total pressure of $\pu{99.652 kPa}$, the distillate contained $\pu{1.27 g}$ of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is $\pu{85.140 kPa}$ at $\pu{95.2°C}$?


Total pressure, $$P_{\text{total}}=\pu{99.652 kPa}\\ P_{\text{water}}=p_B=\pu{85.140 kPa}\\ P_{\text{liquid}}=p_A=\pu{(99.652−85.140) kPa}=14.512 kPa\\ \text{and} \frac{m_a}{m_b}= \frac{1.27}{1}\\ \text{or }= \frac{m_a}{m_b}= \frac{p_AM_A}{p_BM_B}\\ \text{or,} M_A=(\frac{m_A}{m_B})(\frac{p_BM_B}{p_A})\\ M_A=1.27\times (\frac{\pu{85.140kPa} \times \pu{18 g mol−1}}{\pu{14.512 kPa}})\\ M_A≃\pu{134.1 g mol−1}$$

Just can someone explain point 2) intuitively and in basic terms ? This sum and the usage of Raoult's law in steam distillation is killing me here.

(Maybe if someone feels that I need a brush up of steam distillation topic please gladly do so)

still confused here..assistance will be appreciated pls

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    $\begingroup$ A mixture of two immiscible liquids is treated as two seperate phases and thus the total pressure will be sum of vapour pressure of both the liquids. $\endgroup$ – Nisarg Bhavsar Jun 5 at 3:55
  • $\begingroup$ @NisargBhavsar...okay but that was not the required clarification $\endgroup$ – puma Jun 5 at 7:04
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    $\begingroup$ My comment meant that Raoult's law is not applicable for immisible liquids. Both act independently. $\endgroup$ – Nisarg Bhavsar Jun 5 at 7:06
  • $\begingroup$ @NisargBhavsar Then why was the problem solved that way ? $\endgroup$ – puma Jun 5 at 8:08
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    $\begingroup$ You bring us a solution that never mentions Raoult's law, never uses it, and even couldn't possibly have used it. Then you ask to explain why it uses Raoult's law. That's what is going on. It doesn't. That's the explanation. $\endgroup$ – Ivan Neretin Jun 8 at 8:32

Now, first of all does it hold for a mixture of immiscible liquids ?

From Wikipedia on Raoult's law,

it states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. In consequence, the relative lowering of vapor pressure of a dilute solution of nonvolatile solute is equal to the mole fraction of solute in the solution.

Bold mine, now see this remark I have paraphrased from the section of consequences of ideal solution in wiki:

Free energy of mixing is always negative. Ideal solutions are always completely miscible.


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