1. What should be the minutaum distribution coefficient that permits removal of 65% of a solute from 75 mL aqucous solution when extracted with 3 portions of 25 ml.of toluene. A) 0.16 D) 046 C) 1.26 D) 0.92 E) 1.62

To solve this problem, we can use the formula for the distribution coefficient (Kd) in a multi-extraction scenario:<br /><br />\[ K_d = \frac{C_{\text{solvent}}}{C_{\text{solvent}}} \]<br /><br />Where:<br />- \( C_{\text{solvent}} \) is the concentration of the solute in the solvent phase.<br />- \( C_{\text{solvent}} \) is the concentration of the solute in the aqueous phase.<br /><br />Given:<br />- Initial concentration of the solute in the aqueous phase (\( C_{\text{initial}} \)) = 1 (assuming 100% initial concentration for simplicity)<br />- Volume of aqueous phase (\( V_{\text{aq}} \)) = 75 mL<br />- Volume of solvent phase (\( V_{\text{solvent}} \)) = 25 mL<br />- Number of extractions (\( n \)) = 3<br />- Desired removal of 65% of the solute<br /><br />After one extraction, the concentration of the solute in the aqueous phase will be reduced by a factor of \( K_d \). After three extractions, the concentration of the solute in the aqueous phase will be reduced by \( K_d^3 \).<br /><br />The amount of solute remaining in the aqueous phase after three extractions is 35% of the initial amount. Therefore, we can set up the equation:<br /><br />\[ 0.35 = K_d^3 \]<br /><br />Solving for \( K_d \):<br /><br />\[ K_d = \sqrt[3]{0.35} \]<br /><br />\[ K_d \approx 0.146 \]<br /><br />Therefore, the minimum distribution coefficient that permits the removal of 65% of a solute from 75 mL aqueous solution when extracted with 3 portions of 25 mL of solvent is approximately 0.146.<br /><br />So, the correct answer is:<br />A) 0.16