Answer:
5 extractions to extract at least 99.5% of 50.0 g Compound A from 100 mL of water
Step-by-step explanation:
If K=10...
Partition coefficient is defined as the ratio of concentrations of a compound A in two inmiscibles solvents:
K = 10 = Conc. Organic solvent / Conc. Water
Usually organic phase over aqueous phase.
In the first 20mL extraction, the organic solvent will extract:
10 = X / 20mL / (50.0g - X) / 100mL
10 = 100X / (1000-20X)
10000 - 200X = 100X
10000 = 300X
X = 33.3g of compound A are extracted in the first extraction
Remember you want to extract 99.5%, that is 50.0g*99.5% = 49.75g
In the aqueous phase remain: 50-33.3g = 16.7g:
Second extraction:
10 = X / 20mL / (16.7g - X) / 100mL
10 = 100X / (334-20X)
3340 - 200X = 100X
3340 = 300X
11.1g are extracted and will remain: 16.7g - 11.1g = 5.6g
Third extraction:
10 = X / 20mL / (5.6g - X) / 100mL
10 = 100X / (112-20X)
1120 - 200X = 100X
1120 = 300X
3.8g are extracted and will remain: 5.6g - 3.8g = 1.8g
Fourth extraction:
10 = X / 20mL / (1.8g - X) / 100mL
10 = 100X / (36-20X)
360 - 200X = 100X
360 = 300X
1.2g are extracted and will remain: 1.8g -1.2g = 0.6g
Fifth extraction:
10 = X / 20mL / (0.6g - X) / 100mL
10 = 100X / (12-20X)
120 - 200X = 100X
120 = 300X
0.4g are extracted. The total extractions gives:
33.3g + 11.1g + 3.8g + 1.2g + 0.4g = 49.8g
That means, you need to do:
5 extractions to extract at least 99.5% of 50.0 g Compound A from 100 mL of water