The distribution coefficient, k = concentration in hexanes ÷ concentration in water,
between hexanes and water for solute A is 7.5.
(a) What weight of A would be removed from a solution of 10 g of A in 100 mL of water
by a single extraction with 100 mL of hexanes?
(b) What weight of A would be removed by four successive extractions with 25 mL
portions of hexanes?
(c) How much hexanes would be required to remove 98.5% of A in a single extraction?

To solve these problems, we need to use the distribution coefficient, \( k \), which represents the ratio of the concentration of solute \( A \) in hexanes to that in water:

\[
k = \frac{[\text{A}]_{\text{hexanes}}}{[\text{A}]_{\text{water}}} = 7.5
\]

We will solve each part step-by-step.

### (a) Weight of A removed by a single extraction with 100 mL of hexanes

Let's denote:
- \( V_{\text{water}} = 100 \) mL (volume of water)
- \( V_{\text{hexanes}} = 100 \) mL (volume of hexanes)
- \( m_{\text{A,initial}} = 10 \) g (initial amount of A in water)

The amount of solute \( A \) remaining in the water after extraction is given by:

\[
\frac{m_{\text{A, remaining}}}{V_{\text{water}}} = \frac{m_{\text{A,initial}} - m_{\text{A, extracted}}}{V_{\text{water}}}
\]

The concentration ratio between hexanes and water after extraction is:

\[
k = \frac{\frac{m_{\text{A, extracted}}}{V_{\text{hexanes}}}}{\frac{m_{\text{A, remaining}}}{V_{\text{water}}}}
\]

Rearranging for \( m_{\text{A, extracted}} \):

\[
m_{\text{A, extracted}} = \frac{k \cdot V_{\text{hexanes}}}{V_{\text{water}} + k \cdot V_{\text{hexanes}}} \cdot m_{\text{A,initial}}
\]

Substitute the values:

\[
m_{\text{A, extracted}} = \frac{7.5 \times 100}{100 + 7.5 \times 100} \times 10
\]

Let's calculate this.The weight of solute \( A \) removed from the solution by a single extraction with 100 mL of hexanes is approximately **8.82 g**.

### (b) Weight of A removed by four successive extractions with 25 mL portions of hexanes

For multiple extractions, the remaining amount of \( A \) in the water after each extraction is reduced iteratively. The formula for the amount remaining after \( n \) extractions is:

\[
m_{\text{A, remaining after } n \text{ extractions}} = m_{\text{A, initial}} \left( \frac{V_{\text{water}}}{V_{\text{water}} + k \cdot V_{\text{hexanes per extraction}}} \right)^n
\]

Given:
- \( V_{\text{hexanes per extraction}} = 25 \) mL (volume of hexanes per extraction)
- \( n = 4 \) (number of extractions)

The amount extracted after 4 extractions can be found by:

\[
m_{\text{A, extracted after 4 extractions}} = m_{\text{A, initial}} - m_{\text{A, remaining after 4 extractions}}
\]

Let's calculate this.The weight of solute \( A \) removed by four successive extractions with 25 mL portions of hexanes is approximately **9.85 g**.

### (c) Volume of hexanes required to remove 98.5% of A in a single extraction

To determine the volume of hexanes required to remove 98.5% of solute \( A \) in a single extraction, we set up the equation:

\[
\frac{m

Explore how to calculate the removal of solute A from a solution using chemical extraction. This problem involves using the distribution coefficient and mass balance principles to determine weights of A removed with different extraction scenarios.

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