How many 20 mL extractions of organic solvent are needed to extract at least 99.5% of 50.0 g Compound A from 100 mL of water if K

5 extractions to extract at least 99.5% of 50.0 g Compound A from 100 mL of water

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:

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.