At 25 degree C. 1.0 g of sample A will dissolve in 40.0 mL of water, in 5.1 ml.
ID: 495652 • Letter: A
Question
At 25 degree C. 1.0 g of sample A will dissolve in 40.0 mL of water, in 5.1 ml. of Chloroform, in 80.0 ml. of benzene, or in 13.7 ml. of diethyl ether. Calculate the solubility (in grams/100 mL.) of sample A in these four solvents. Which of these three organic solvents would be the best choice for extracting sample A from an aqueous solution and why? Calculate the mole fraction of each of the compound in the following 50 g water and 50 g of acetone (C_3H_6O) 12.5 g ethanol (C_2H_6O), 30.0 g methanol (CH_4O) and 20.0 g 2-propanol (C_3H_5O) What are the partial pressures of water and acetone in questions 7a if the total pressure of tin- mixture is 760 mmHg?Explanation / Answer
5. Solubility of compound A in g/100 ml
solvent : water
solubility = 1 g x 100 ml/40 ml = 2.5 g/100 ml H2O
solvent : chloroform
solubility = 1 g x 100 ml/5.1 ml = 19.61 g/100 ml CHCl3
solvent : benzene
solubility = 1 g x 100 ml/80 ml = 1.25 g/100 ml benzene
solvent : diethylether
solubility = 1 g x 100 ml/13.7 ml = 7.3 g/100 ml diethylether
So maximum solubility for the compoind is in chloroform (CHCl3) and thus this is the best solvent for the extraction of compound A from aqueous medium.
6. Mole fraction
1. moles H2O = 50 g/18 g/mol = 2.8 mol
moles acetone = 50 g/58.08 g/mol = 0.86 mol
total moles = 3.66 mol
mole fraction H2O = 2.8/3.66 = 0.765
mole fraction acetone = 1-0.765 = 0.235
b. moles EtOH = 12.5 g/46.07 g/mol = 0.27 mol
moles MeOH = 30 g/32.04 g/mol = 0.94 mol
moles iPrOH = 20 g/60.1 g/mol = 0.33 mol
Total moles = 1.54 mol
mole fraction EtOH = 0.27/1.54 = 0.176
mole fraction MeOH = 0.94/1.54 = 0.610
mole fraction iPrOH = 0.33/1.54 = 0.214
c. Partial pressure of water = 0.765 x 760 = 581.4 mmHg
Partial pressure of acetone = 0.235 x 760 = 178.6 mmHg
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