4. A 50.0 mL aqueous solution contains 1.00 g of compound X. Calculate the mass of compound X extracted into ethyl acetate for each of the following extraction procedures if the partition coefficient for the extraction is 5.00, with X being more soluble in ethyl acetate. Show all work for full credit.

(A). 0.5g, (B). 0.83 g and (C). 0.875 g.

Explanation:

So, we are given the following data or parameters which are going aid us in solving this particular Question/problem.

=> Volume of X given = 50.0 mL = 50 × 10^-3 L and the mass of the aqueous solution contains = 1.00 g of compound X.

So, let us dive straight into the solution to this question;

(A). For a single 10.0 mL extraction with ethyl acetate, we have ;

(W1/V) ÷ (W- W1)/Vn = Kd.

(W1/50) ÷ (1 - w1)/10 = 1/5.

Thus, solving for w1, we have that w1 = 0.5.

Amount extracted = 1 - 0.5 = 0.5g.

(B). For a single 50.0 mL extraction with ethyl acetate, we have;

(W1/50) ÷ (1 - w1)/50 = 1/5.

W1 = 0.17.

Amount extracted = 1 - 0.17 = 0.83g.

(C). For Three 10.0 mL extractions with ethyl acetate, we have;

{(1/5 × 50)÷ (1/5 × 50) + 10}^3 × 1 = 0.125

The amount extracted = 1 - 0.125 = 0.875g.

c, liquids evaporate, creating a gas