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: Where:- is the concentration of the solute in the solvent phase.- is the concentration of the solute in the aqueous phase.Given:- Initial concentration of the solute in the aqueous phase ( ) = 1 (assuming 100% initial concentration for simplicity)- Volume of aqueous phase ( ) = 75 mL- Volume of solvent phase ( ) = 25 mL- Number of extractions ( ) = 3- Desired removal of 65% of the soluteAfter one extraction, the concentration of the solute in the aqueous phase will be reduced by a factor of . After three extractions, the concentration of the solute in the aqueous phase will be reduced by .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: Solving for : 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.So, the correct answer is:A) 0.16