Respuesta :
For the extraction of at least 95% of Y in water, 3 extractions are to be performed.
Distribution coefficient can be defined as the ratio of the concentration of solute in an organic solvent to water.
Distribution coefficient = [tex]\rm \dfrac{concentration\;in\;solvent}{concentration\;in\;water}[/tex]
- For the first extraction, the concentration in organic solvent = X per 10 ml
Concentration in water = 100-X per 50 ml
Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]
4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]
4 = [tex]\rm \dfrac{50X}{1000-X}[/tex]
4000 - 4X = 50X
X = 74.1 %
Thus, after the first extraction, the amount of Y extracted is 74.1%.
We have to extract at least 95% of Y. Thus, the second extraction is performed.
- Remaining y = 100 - 74.1
Remaining y = 25.9%
The concentration in organic solvent = X per 10 ml
Concentration in water = 25.9 -X
Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]
4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]
4 = [tex]\rm \dfrac{50X}{259-X}[/tex]
1036 - 4X = 50X
X = 19.2%
Thus, after the second extraction the amount of Y extarcted = first extraction + second extraction
The amount of Y extracted = 74.1 + 19.2 %
The amount of Y extracted = 93.3%
- To reach at least 95% extraction, the third extraction has to be performed.
The remaining Y for third extraction = 100 - 93.3
The remaining Y for the third extraction = 6.7%
Concentration in water = 100 - 6.7
Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]
4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]
4 = [tex]\rm \dfrac{50X}{67-X}[/tex]
268 - 4X = 50X
X = 5.0%
The total extraction after third extraction = 93.3 + 5%
The total extraction after third extraction = 98.3%.
Thus for the extraction of at least 95% of Y in water, 3 extractions are to be performed.
For more information about the distribution coefficient, refer to the link:
https://brainly.com/question/13599982
