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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 $\pu{95.2°C}$?

Solution:

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. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ Jun 8 at 8:32
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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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I believe this excerpt from sciencedirect may answer your question: "Vapor Pressures and Boiling Points in a Mixture of Immiscible Liquids

When two immiscible liquids are present in a mixture, the total vapor pressure of the mixture is equal to the sum of each liquid's vapor pressure, independently of their concentration, as long as there is enough of each liquid so they can both reach equilibrium. The condition to reach this state is that the mixture must be in constant agitation. If the mixture is let to rest, both liquids will form layers and the vapor pressure of the mixture will equal the vapor pressure of the liquid forming the top layer. The bottom layer will not have any vapor pressure as it is in contact with the top layer, thus preventing any evaporation. Therefore, it is necessary for both liquids to be in constant agitation to let the phenomenon of added vapor pressures to take place.

Consequently, when two immiscible liquids are in a mixture, their boiling point can be greatly decreased. For example, the boiling point (vapor pressure of 101,325 Pa at standard pressure) of water is 100°C, and the boiling point of m-xylene is 138.7°C [3]. When these two immiscible liquids are present in a mixture, their vapor pressures add up to make the total vapor pressure of the mixture. The boiling point of the mixture is still at 101,325 Pa since it is operating at standard pressure, however the temperature required to reach the boiling point is now below 100°C. As a matter of fact, at 93°C the vapor pressures of water and m-xylene are 78,494 Pa and 24,431 Pa, respectively [3, 4]. This adds up to a total vapor pressure of 102,925 Pa, sufficient for the solution to boil. This is the principle on which steam distillation operates." https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/vapor-pressure

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