A 25.0 mL aqueous solution contains 2.50 g of compound Λ. Calculate the mass of compound Λ in a single 5.0 mL extraction with ethyl acetateif the partition coefficient (CEtOAc/Cwater) for the extraction is 1.5 (Λ being more soluble in ethyl acetate)

Answer 1

To solve this problem, we can use the partition coefficient equation:

Partition coefficient = (mass of compound Λ in ethyl acetate) / (mass of compound Λ in water)

Given that the partition coefficient (CEtOAc/Cwater) is 1.5, we can set up the equation:

1.5 = (mass of compound Λ in ethyl acetate) / (mass of compound Λ in water)

We know that the initial aqueous solution contains 2.50 g of compound Λ in 25.0 mL. To calculate the mass of compound Λ in a 5.0 mL extraction, we need to use proportions.

2.50 g / 25.0 mL = x g / 5.0 mL

Cross-multiplying, we get:

25.0 mL * x g = 2.50 g * 5.0 mL
25.0x = 12.5

Solving for x, the mass of compound Λ in a 5.0 mL extraction, we get:

x = 12.5 / 25.0
x = 0.5 g

Now, we can substitute the values back into the partition coefficient equation:

1.5 = (mass of compound Λ in ethyl acetate) / 0.5 g

Cross-multiplying, we get:

1.5 * 0.5 g = mass of compound Λ in ethyl acetate
0.75 g = mass of compound Λ in ethyl acetate

Therefore, the mass of compound Λ in a single 5.0 mL extraction with ethyl acetate is 0.75 g.