Respuesta :
Explanation:
The given data is as follows.
Partition coefficient, K = 6.2
Volume of phase 1, [tex]V_{1}[/tex] = 74.0 mL
Volume of phase 2, [tex]V_{2}[/tex] = 17.0 mL
So, after one extraction fraction of solute remaining is given as follows.
q = [tex]\frac{V_{1}}{V_{1} + KV_{2}}[/tex]
After 3 times extraction, fraction of S remaining is as follows.
q = [tex][\frac{V_{1}}{V_{1} + KV_{2}}]^{3}[/tex]
Putting the given values into the above formula as follows.
q = [tex][\frac{V_{1}}{V_{1} + KV_{2}}]^{3}[/tex]
= [tex][\frac{74.0 ml}{74.0 ml + 6.2 \times 17.0 ml}]^{3}[/tex]
= [tex][\frac{74.0 ml}{179.4 ml}]^{3}[/tex]
= 0.0699
Thus, we can conclude that the fraction of S remaining in the aqueous phase is 0.0699.
