A student had 150. mL of an aqueous solution that contains 15.0 grams of malonon

Question

A student had 150. mL of an aqueous solution that contains 15.0 grams of malononitrile. CH2(CN)2. The student wanted to extract the malononitrile with ether. At room temperature the solubility of malononitrile in ether is 20.0g/ 100 mL and in water is 13.3g/ 100 mL.

Calculate the total mass of malononitrile that can be extracted with three successive 50mL portions of ether. Show your work.

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Explanation / Answer

STep 1:

Partition coefficient (Kd) = [Solute]o / [Solute]aq =

Kd = (20.0 g/100mL) / (13.3 g/100mL) = 1.503

Step2:

With first 50 mL portion of ether,

1.503 = (x /50)ether / (15-x)/100 water

1.503 = 2x / 15-x

22.54 - 1.503x = 2x

3.503 x = 22.54

x = 6.43 g

So, 6.43 g of malanonitrile is extracted with fist 50 mL portion.

Thus, portion extracted = 6.43 / 15 = 0.428 portion of total is extracted.

Total fraction extracted = 1-(1-x)^N

here N = 3

x = 0.428

1-(1-x)^N = 0.813

In 15 g of mass = 15 x 0.813 = 12.2 g of malanonitrile

With three portions

1-x = 1-1.47 =

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