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In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 10 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2 + 20 mL H2O a. How many moles of acetone were in the reaction mixture? Remember M = mol/L and M x vol. = mol b. What was the molarity of acetone in the reaction mixture? c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50.0 mL and keeping the concentrations of H+ ion and I2 as in the original mixture?
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