Answer:
At least 3 three extractions are required to recover at least 99.5 % of the material in the organic layer.
Explanation:
The partition coefficient of a solute (S) soluble in two immiscible solvents is given by the following formula
[tex]K_S=\frac{[S]_2}{[S]_1}[/tex]
The lower layer is taken as an aqueous layer (1), while the upper layer is organic (2). The fraction of solute remaining in the aqueous layer is given by the following formula
[tex]q^n=(\frac{V_1}{V_1 \times KV_2})^n[/tex]
Here, n denotes the number of extraction, and q^n represents the fraction of solute remaining in aqueous solvent after n number of extraction. According to the given data, the fraction of solute remaining in the aqueous layer after multiple extractions is 0.005, i.e., q^n=0.005. Mathematically,
[tex]0.005=(\frac{50}{50 + 50\times10})^n\\\\0.005=(0.091)^n[/tex]
Taking log on both sides
[tex]log(0.005)=nlog(0.091)[/tex]
[tex]log(0.005)=nlog(0.091)\\\\-2.301=-n1.04\\n=2.21[/tex]
The above calculations show that the number of extractions should be greater than 2, i.e, at least 3, in order to achieve extraction greater than 99.5 %.