Question

Liquids B and C are partially miscible at 25 oC. When one starts with 1 mol of C at 25 oC and isothermally adds B a little at a time, a two-phase system first appears when a bit more than 0.125 mol of B has been added; continuing to add B, one finds the two-phase system becomes a one-phase system when a total of 3 mol of B has been added. For a system consisting of 2.5 mol of B and 2 mol of C at 25 oC, find the number of moles of B and of C present in each phase.

Answer 1

Answer:

$0.8\overline 3$ moles of A and 2.5 moles of B in the one-phase system

$1.1 \overline 6$ moles of A and $0.1458 \overline 3$ moles of B in the two-phase system

$0.3541 \overline 6$ moles of B remains in the system

Explanation:

The given parameters are;

The extent of miscibility of liquid B and C = Partially miscible

The number of moles of liquid B added to 1 mole of liquid A that forms a two-phase system = 0.125 mol of liquid B

The number of moles of liquid B added to 1 mole of liquid A that forms a one -phase system = 3 moles of liquid B

Whereby the system consist of 2.5 mol of B and 2 mol of C at 25°C, we have;

The 1 mole of A mixes with 3 moles of B to form a single phase solution

Therefore;

2.5 moles of B will mix with (1/3)×2.5 moles of A = 5/6 moles of A

The remaining number of moles of A in the system = 2 - 5/6 =  7/6 moles of A

Similarly, we have;

At least, 0.125 mole of B combines with 1 mole of A to form a two-phase system

7/6 moles of A will combine with 7/6 × 0.125 = 7/48 moles of B to form a two-phase system

The number of moles of B left = 0.5 - 7/48 = 17/48 = 0.3541$\overline 6$ moles of A

Therefore, we have;

5/6 moles of A and 2.5 moles of B in the one-phase system

7/6 moles of A and 7/48 moles of B in the two-phase system

17/48 moles of B remaining in the system