Respuesta :
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.
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. Â
Thus, the ration of the solubility is, Â
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, Â
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. Â
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, Â
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. Â