To solve this problem, let's carefully compute the fraction of solute $A$ remaining in each case.

Partition Coefficient Equation:

The partition coefficient $K$ is defined as: $K = \frac{C_{\text{heptane}}}{C_{\text{water}}}$ where $C_{\text{heptane}}$ and $C_{\text{water}}$ are the concentrations of solute $A$ in the heptane and water layers, respectively.

The formula for the fraction of solute remaining in the aqueous phase after one extraction is: $f = \frac{1}{1 + K \cdot \frac{V_{\text{heptane}}}{V_{\text{water}}}}$ where:

• $K = 2.8$ (given),
• $V_{\text{heptane}}$ is the volume of heptane used,
• $V_{\text{water}} = 50 \, \text{mL}$.

For multiple extractions, the fraction remaining after $n$ extractions is given by: $f_n = f^n$ where $f$ is the fraction remaining after a single extraction.

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Case 1: Single Extraction with 400 mL of Heptane

Here, $V_{\text{heptane}} = 400 \, \text{mL}$. Using the single-extraction formula: $f = \frac{1}{1 + 2.8 \cdot \frac{400}{50}}$ Simplify: $f = \frac{1}{1 + 2.8 \cdot 8} = \frac{1}{1 + 22.4} = \frac{1}{23.4}$ $f = 0.0427$ Thus, the fraction of $A$ remaining is: $f \cdot 100 = 4.27\%$

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Case 2: Four Extractions with 100 mL of Heptane Each

For each extraction, $V_{\text{heptane}} = 100 \, \text{mL}$. Using the same formula: $f = \frac{1}{1 + 2.8 \cdot \frac{100}{50}}$ Simplify: $f = \frac{1}{1 + 2.8 \cdot 2} = \frac{1}{1 + 5.6} = \frac{1}{6.6}$ $f = 0.1515$ Thus, the fraction remaining after one extraction is $15.15\%$.

For four extractions, the total fraction remaining is: $f_4 = f^4 = (0.1515)^4$ Simplify: $f_4 = 0.000525$ Convert to a percentage: $f_4 \cdot 100 = 0.0525\%$

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Final Results:

1. Fraction of $A$ remaining after one extraction with 400 mL: $4.27\%$
2. Fraction of $A$ remaining after four extractions with 100 mL each: $0.0525\%$

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Conclusion:

Using four extractions of 100 mL is significantly more effective than a single extraction with 400 mL.

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Would you like further clarification or additional breakdowns? Here are 5 related questions to expand your understanding:

1. What happens to the effectiveness of extractions as the partition coefficient increases?
2. How would the results change if the volumes of water and heptane were equal?
3. Can this method be extended to other solutes with lower partition coefficients?
4. What are the practical limitations of performing multiple extractions in a lab setting?
5. How is the fraction remaining influenced by changing the total volume of heptane?

Tip: Always use multiple smaller extractions rather than one large extraction to maximize solute transfer efficiency!