Fun With Fluids!

Relevant wiki: Vapor Pressure and Raoult's Law

At 1st, I will like to try to clear a doubt which generally many people are having - When the equilibrium is reached , some part of liquid will be evaporated , so , the composition in the solution must change ? ( As asked to me by @Prakhar Bindal )

I thought on it a bit and got one line that these laws are valid at a low pressure and high temperature condition (Ideal gas condition ) , so we can say that since the number of moles evaporated is very-very less , there is no considerable change in the composition of the solution.

Also , I am mentioning the Ideal gas condition in the assumption.

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I will solve by considering ratios as mole fraction is all about the ratio of moles and Raoult's law is all about mole fraction!
Here $n$ represents mole & $P$ represents pressure ( Here partial pressure )
Beaker $\mathbb I$ :
At equilibrium,
In solution ${n}_{A} : {n}_{B} = 1 : 4$
In Vapour phase ${P}_{A} : {P}_{B} = 100 \times \frac {1}{5} : 200 \times \frac {4}{5} = 1 : 8$

Beaker $\mathbb{ II}$
After adding C and at equilibrium
In solution ${n}_{A} : {n}_{B} : {n}_{C} = 1 : 8 : 1$
In Vapour phase ${P}_{A} : {P}_{B} : {P}_{C} = 1 : 16 : 3$

Beaker $\mathbb{ III}$
After adding D and at equilibrium
In solution ${n}_{A} : {n}_{B} : {n}_{C} : {n}_{D} = 1 : 16 : 3 : 16$
In Vapour phase ${P}_{A} : {P}_{B} : {P}_{C} : {P}_{D} = 1 : 32 : 9 : 64$

Beaker $\mathbb{ IV}$
At equilibrium,
In solution ${n}_{A} : {n}_{B} : {n}_{C} : {n}_{D} = 1 : 32 : 9 : 64$
Total pressure
$P = {P^\circ}_{A} {\chi }_{A} + {P^\circ}_{B} {\chi}_{B} + {P^\circ}_{C} {\chi}_{C} + {P^\circ}_{D} {\chi}_{D} = \frac {17400}{53}$

(Where , ${\chi}_{i}$ represents the mole fraction of ${i}^\text{th}$ component of liquid in solution )

$\boxed { M\times N = 922200 }$ .