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------------------------------------------------ Two hundred kilograms of a 40% solution of A in B (free solvent) is equilibrated with 170 kg of solvent S containing 2% of A. At equilibrium, the raffinate phase has a mass of 137 kg and contains 68% B and 9% S.

What is the selectivity or separation factor of A?

The selectivity or separation factor of A in the given system is calculated by balancing the amount of A in the beginning with the amount of A in the extract and raffinate phases. It comes out to be 1.39.

Explanation:

The selectivity factor or separation factor for a particular system is defined as the ratio of the weight fraction of the solute in the extract phase to the weight fraction of the solute in the raffinate phase.

In this case, we first calculate the amount of solute A in the two phases.

The amount of component A in the 40% mixture is 200 kg * 40/100 = 80 kg.

The amount of A in the solvent S is 170 kg * 2/100 = 3.4 kg

Hence, the total amount of A in the system initially is 80 kg + 3.4 kg = 83.4 kg.

Now, from the raffinate data, we calculate the amount of A left. If the raffinate weighs 137 kg and has 68% B, 9% S, it also has 23% A.

Hence, the amount of A in the raffinate is 137 kg * 23/100 = 31.51 kg.

With mass balance, we conclude that the amount of A in the extract phase is 83.4 kg - 31.51 kg = 51.89 kg.

The separation factor or selectivity of A thus becomes the weight fraction of A in the extract phase divided by the weight fraction of A in the raffinate phase, which equals (51.89 kg / 200 kg)/(31.51 kg / 170 kg) = 1.39.

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