A 200g mixture containing 950 by weight of A and 50 of B is

A 20.0g mixture containing 95.0% (by weight) of A and 5.0% of B is recrystallized in toluene (Boiling Point: 110 degrees C). What amount of solvent is needed to obtain pure A, and how much pure A will be recovered? Please show calculation and provide an explanation of result. A) 20 degrees C Solubility: 1.5 g/ 100 mL 110 degrees C Solubility: 10g/ 100 mL B) 20 degrees C Solubility: 0.50g/ 100 mL 110 degrees C Solubility: 8.0g/ 100 mL

Solution

Solution:

Here volume is 100 mL.Here 1.5 g of A will dissolve in 100 mL of toluene at 20 degree celsius.

In 20 g sample contains 95% A is

20*95/100=95/5=19 grams A

In 20 g sample contains 5% B is

20*5/100=5/5=1 gram B.

Toluene need to dissolve 19 g A is

100 mL *(19/10)=190 mL

This means that 190 mL toluene need to dissolve 19 g A at 110 C degree celsius.

This 190 mL toluene dissolves all of the B because B has 1 g and the solubility of B at 110 degree celsius is 8 g.

When we cool 190 mL toluene at 20 degree celsius A comes out is

1.5 g* (190/100)=2.85 g A dissolve.Thus 19-2.85=16.15 g recrystallize.So the percentage recovery is (16.15/20)*100=80.75%