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