Answer:
a. X = 0,909
b. X = 0,965
c X = 0,997
Explanation:
The partition coefficient (k) is defined as:
k = Solute in ether / solute in water
a. [tex]3,34 = \frac{\frac{X}{300mL} }{\frac{1-X}{100mL} }[/tex]
Where X is the fraction of solute extracted
3,34 = X / 3-3X
10,02-10,02X = X
10,02 = 11,02X
X = 0,909
b. First extraction:
[tex]3,34 = \frac{\frac{X}{100mL} }{\frac{1-X}{100mL} }[/tex]
3,34 = X / 1-X
3,34 - 3,34X = X
3,34 = 4,34X
X = 0,770
That means solute in water will be: 1-0,770 = 0,23
Second extraction:
The second extraction will extract the same fraction of solute, as now you have 0,230 of solute in water you will extract:
0,230×0,770 = 0,177
Third extraction:
In the same way, the third extraction will extract:
(0,230-0,177)×0,770 = 0,018
Fraction in water×Fraction extracted
That means total solute extracted is:
0,770 + 0,177 + 0,018 = 0,965
c. Extracting with 50mL of ether:
First extraction
[tex]3,34 = \frac{\frac{X}{50mL} }{\frac{1-X}{100mL} }[/tex]
3,34 = 2X / 1-X
3,34 - 3,34X = 2X
3,34 = 5,34X
X = 0,625
Second extraction:
(1-0,625)×0,625= 0,234
Third Extraction:
(1-0,625-0,234)×0,625= 0,088
Fourth extraction:
(1-0,625-0,234-0,088)×0,625= 0,033
Fifth extraction:
(1-0,625-0,234-0,088-0,033)×0,625= 0,013
Sixth extraction:
(1-0,625-0,234-0,088-0,033-0,013)×0,625= 0,004
Total extractions gives:
0,625+0,234+0,088+0,033+0,013+0,004 = 0,997
I hope it helps!
