Compound A has a partition coefficient (K) of 7 when comparing its solubility in CH2Cl2 to water ( K=11, [solubility of A in g/ml in CH2Cl2] divided by [solubility of A in g/ml in H2O]). If we take 17.0 g of A and partition it thoroughly between 150 ml of CH2Cl2 and 100 ml of water, what is the equation which will tell us how much of A (which is represented by x) is in the water layer after this partitioning ?

The equation which will tell us how much of A that is mis inmthe water layer after partitioning is: 7 = (17 - x) g / 150 mL ÷ x g /100 mL

Explanation:

A partition coefficient is the ratio of the concentration of a substance in one solvent phase to the concentration in a second solvent phase when the two concentrations are at equilibrium. Usually the two phases are an organic phase and an aqueous phase. Thus, the partition coefficient K, of a compound is the ratio of the compound's concentration in the organic layer compared to the aqueous layer.

K = C₁/C₂ at equilibrium

In the compound A given, CH₂Cl₂ is the organic phase while water is the aqueous phase

Amount of A that is partitioned in between dichloromethane, CH₂Cl₂ and water, H₂O is 17.0 g

Let the amount of A that is dissolved in water be x g

Solubility of A in water given in g/mL = (x / 100) g/ml

Amount of A dissolved in dichloromethane, CH₂Cl₂ = (17 - x) g

Solubility of A in dichloromethane, CH₂Cl₂ given in g/mL = (17 - x/150) g/mL

Since the partition coefficient, K of compound A when comparing its solubility in CH₂Cl₂ to water is 7, that is;

K = [solubility of A in g/ml in CH₂Cl₂] / [solubility of A in g/ml in H2O] = 7

The equation for the amount of A in the water layer is given as follows:

7 = (17 - x) g / 150 mL ÷ x g /100 mL

Solving for x

7 = (17 -x) × 100 / 150x

7 × 150x = (17 - x) × 100

1050x = 1700 - 100x

1150x = 1700

x = 1700/1150

x = 1.48 g

i don,t know so just answer mine

the answer is 75.0%.