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katen-ka-za [31]
3 years ago
6

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

ion using 100 mL of ethyl acetate per extraction if the partition coefficient is 8
Chemistry
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

<em></em>

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