The given question is not complete, the complete question is:
Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100mL. The ratio of these quantities is equal to the partition coefficient, k, which equals What weight of malononitrile would be recovered by extraction of (a) three 100-mL portions of ether and (b) one 300-mL portion of ether? SHOW WORK (Can be written in pen and attached to report). Suggestion: For each extraction, let x equal the weight extracted into the ether layer. In part (a), for the first of the three extractions, the concentration of malononitrile in the ether layer is x/100 and in the water layer is (30-x)/100.
Answer:
The correct answer is 10 grams and 18 grams.
Step-by-step explanation:
Based on the given question, 20 gram per 100 ml is the solubility of malononitrile in ether, and 13.3 gram per 100 ml is the solubility of malononitrile in water.
Thus, the ration of the solubility is,
Solubility in water/solubility in ether = 20/13.3 = 1.50
a) Let w be the weight of malononitrile extracted into water in every extraction. Then the concentration of the ether layer will be w/100. The concentration in the water layer will be 30-w/300. Now the ratio will be,
Ratio = w/100 / (30-w)/300
1.50 = w/100 * 300 (30-w)
w = 10
Hence, the weight of malononitrile recovered by extraction is 10 grams.
b) The concentration in the ether layer will be w/300. The concentration in the water layer will be (30-w) / 300. Now the ratio will be,
Ratio = w/300 / (30-w) / 300
1.50 = w/300 * 300 (30-w)
w = 18
Hence, 18 grams is the weight of malononitrile recovered by extraction.