Question

If the partition coefficient of CBN from water into ethanol is 4, what fraction "a" of the CBN would remain in a 448 mL aqueous sample if it was extracted with 81 mL of ethanol twice?

a) 1/16
b) 1/8
c) 1/4
d) 1/2

Answer 1

The fraction of CBN remaining in the aqueous solution after two extractions with ethanol is 1/25. This result is obtained by applying the partition coefficient to calculate the remaining fraction after each extraction.

Step-by-step explanation:

To determine the fraction of CBN that would remain in an aqueous sample after being extracted twice with ethanol, we can use the concept of the partition coefficient. The partition coefficient (K) is the ratio of concentrations of a compound in a mixture of two immiscible solvents at equilibrium. Given that K for CBN is 4, we can set up the following equation to find the remaining fraction:

Initial fraction in water = 1 (100%)
First extraction: Fraction remaining in water = 1 / (1 + K) = 1 / (1 + 4) = 1/5
Second extraction: Fraction remaining in water = (1/5) / (1 + K) = (1/5) / (1 + 4) = 1/25

The answer is found by multiplying the fraction remaining after the first extraction by the fraction remaining after the second extraction:

(1/5) × (1/5) = 1/25

Therefore, the fraction 'a' of CBN remaining in the aqueous sample after two extractions with ethanol is 1/25, which is not listed among provided options a, b, c, d. Hence, there might be a mistake in the question or provided options.