Answer:
(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.
