Question

A mixture of 100g of water and 80g of phenol separates into two layers at 50 oC. One layer, L1, consists of 75.4% of water by mole number; the other , L2, contains 97.7% of water by mole number. How many moles of each constituents are in each phase?

Answer 1

Number of moles of water = weight/ molecular weight = 100 g/18 = 5.5 moles of water

Number of moles of phenol = 80/94.11 = 0.85 moles.

Lets consider x1 moles of water and y1 moles of phenol is present in L1.

Since L1 contains 75.4% of water x1 / y1 = 3.06 => y1 = x1/3.06

Similarly,

Lets consider x2 moles of water and y2 moles of phenol is present in L1.

Since L2 contains 75.4% of water x2 / y2 = 29.60  => y2 = x2/29.60

And also we have x1 + x2 = 5.5 mole .. eq (1)

and y1 + y2 = 0.85 moles  .. eq (2)

So substituting y1 and y2 in eq (2), x1/3.06 + x2/29.60 = 0.85  .. eq (3)

Solving  eq (1) and  eq (3) gives x1 = 2.27 moles and x2 =3.22 moles

Also,  y1 = x1/3.06 = 2.27/30.6 = 0.741 moles

y2 =y2 = x2/29.066 = 2.27/29.60 =0.108 moles

So in L1, 2.27 moles of water and 0.741 moles of phenol are present

And, in L2, 3.32 moles of water and 0.108 moles of phenol are present