Respuesta :
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) = [tex]\frac{C_1}{C_2}[/tex]
where; [tex]C_1=[/tex] solubility of compound in the organic solvent and [tex]C_2[/tex] = solubility in aqueous water.
So; we can represent our data as:
[tex]K=(\frac{A_{(g)}}{60mL} )[/tex] ÷ [tex](\frac{B_{(mg)}}{100mL} )[/tex]
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=[tex](\frac{0.6-B(mg)}{60mL} )[/tex] ÷ [tex](\frac{B_{(mg)}}{100mL})[/tex]
4.6 = [tex]\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}[/tex]
4.6 × [tex](\frac{B_{(mg)}}{100mL})[/tex] = [tex](\frac{0.6-B(mg)}{60mL} )[/tex]
4.6 B [tex]*\frac{60}{100}[/tex] = 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B = [tex]\frac{0.6}{3.76}[/tex]
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.
The required caffeine [tex]159 \ gm[/tex] is extracted in [tex]60 ml[/tex] dichloromethane.
Solution:
A solution is a type of homogeneous mixture that is made up of two or more substances. A homogeneous mixture is a type of mixture with a uniform composition. This means that the substances cannot be distinguished easily from one another. Some examples of solutions are salt water, rubbing alcohol, and sugar dissolved in water.
[tex]m=600 mg[/tex] of caffeine
[tex]v=100 ml[/tex] water.
Partition coefficient[tex]=4.6[/tex]
a. Caffeine extracted
[tex]v=100ml[/tex] of water
one portion of [tex]60ml[/tex] of dichloromethane
[tex]K=\frac{C_{1} }{C_{2} }[/tex]
K=partial coefficient
[tex]C_{1} =[/tex]solubility of your compound in an organic solvent
[tex]C_{2}=[/tex]solubility in water(aqueous)
[tex]K=\frac{X g\div 60ml}{Yg\div 100ml}[/tex]
[tex]4.6=\frac{(0.6-Ymg\div 60ml)}{(Ymg\div 100ml)} X+Y=0.6, \ X=0.6-Y[/tex]
[tex]4.6(Ymg\div 100ml)=(0.6-Ymg\div ml)[/tex]
[tex]4.6Y.60\div 100=0.6-Ymg[/tex]
[tex]2.76Y=0.6-Y[/tex]
[tex]2.76Y+Y=0.6\\3.76Y=0.6\\Y=0.6\div 3.76\\Y=0.159gm[/tex]
[tex]Y=159mg[/tex], extracted in 60ml dichloromethane.
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