1. In a reaction involving the iodination of acetone, the following volumes were

Question

1. 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 H20 a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A=Ma * V, where Ma is the molarity of A and V is the volume in liters of the solution of A that was used. __________ moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL , 0.050 l, and the number of moles of acetone was found in Part (a). Again, Ma= no. moles A / V of soln. in liters. ___________ M acetone c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H+ ion and I2 as in the original mixture?

Explanation / Answer

a) Use 20 ml acetone instead of 10 ml and reduce water volume to 10 ml b) The initial concentration of I2 is 10 *10^-3 L* 0.0050 mol/L/ 50 *10^-3 L = 0.001 mol/L. // 1mL = 10^-3 L Therefore rate = 0.001 / 250 mol/(L s) = 4*10^-6 mol/(L s)

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