Two liquids 'A' (molecular mass=40) and B'=(Molecular mass=20) are partially miscible.When 1 mol of A and 3 mol of B are shaken together and allowed to settle, two layer `L_1 and L_2` are formed as shown in diagram.(Mol)(P)[T]

Layer `'L_1'` contains 0.1 mole fraction of 'A' and layer `'L_2'` contains 0.4 mole fraction of A calculate simple ratio of masses of layer `L_1` to layer `L_2`.If your answer is `x/y` then report as `(x+y)`

Answer

25

Explanation

Let. Moles of A in `L_1=x`
Let. Moles of B in `L_1=y`
`:. x/(x+y)=0.1 so 9x=y`…(i)
In `L_2` mole of A =1-x
In `L_2` mole of B=3-y
`:.(1-x)/((1-x)+(3-y))=0.4`
So 1-x=1.6-0.4x-0.4 y
0.4y-0.6x=0.6 …(ii)
by eqn. (i) and (ii) 3x=0.6 and x=0.2 & y=1.8
Mass of `L_1=0.2xx40+1.8xx20=44 g`
Mass of `L_2=0.8xx40+1.2xx20=56 g`
Mass Ratio `L_1` to `L_2` =`44/55=11/14`