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