Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) = ![\frac{C_1}{C_2}](https://tex.z-dn.net/?f=%5Cfrac%7BC_1%7D%7BC_2%7D)
where;
solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷ ![(\frac{B_{(mg)}}{100mL} )](https://tex.z-dn.net/?f=%28%5Cfrac%7BB_%7B%28mg%29%7D%7D%7B100mL%7D%20%29)
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6=
÷ ![(\frac{B_{(mg)}}{100mL})](https://tex.z-dn.net/?f=%28%5Cfrac%7BB_%7B%28mg%29%7D%7D%7B100mL%7D%29)
4.6 = ![\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}](https://tex.z-dn.net/?f=%5Cfrac%7B%28%5Cfrac%7B0.6-B%28mg%29%7D%7B60mL%7D%20%29%7D%7B%28%5Cfrac%7BB_%7B%28mg%29%7D%7D%7B100mL%7D%29%7D)
4.6 ×
=
4.6 B
= 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B = ![\frac{0.6}{3.76}](https://tex.z-dn.net/?f=%5Cfrac%7B0.6%7D%7B3.76%7D)
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.
Carbon dioxide. <span>Cellular respiration is this process in which oxygen and glucose are used to create ATP, carbon dioxide, and water. ATP, carbon dioxide, and water are all products of this process because they are what is created. Carbon dioxide is released as a gas when you exhale.</span>