Answer:
Extraction should be carried out at least 6 times
Explanation:
Distribution coefficient is given according to the following equation
[tex]K=\frac{[S]_2}{[S]_1}[/tex]
Here, S is the solute, and the subscripts denote two different immiscible solvents. The aqueous layer concentration is kept in the denominator, while the organic layer is kept in the numerator.
The percent extraction can be calculated from the following formula:
[tex]q^{n}=(\frac{V_{1}}{V_{1}+KV_{2}})^{n}[/tex]
In the above formula, q is the fraction of the solute remaining in the aqueous layer. V is the volume of solvents and K is the distribution coefficient.
According to the given data, let us consider the volume of both solvents is 1 unit, and for 90% extraction, the remaining fraction in the aqueous solvent layer must be 0.1 (i.e. q∧(n) = 0.1). Placing the data in the equation we get,
[tex]0.1=(\frac{1}{1+(0.5)1})^{n}\\ \\0.1=0.666^{n}[/tex]
Taking the logarithm of the base 10 on both sides, we get,
[tex]log(0.1)=n.log(0.666)[/tex]
[tex]n=\frac{log(0.1)}{log(0.666)} \\ \\ n=5.66[/tex]
The above calculations show that the extraction should be carried out at least six times to achieve extraction greater than 90%.
