Question

1. In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:

5mL 4.0 M acteon + 10ml 1.0 M HCl + 10 mL 0.0050 M I2 + 25 mL H2O

a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, 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 of acetone

b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50mL, 0.050L, and the number of moles of acetone was found in Part (A)

MA = moles of A / V of soln. in L

_____ M acetone

c. How could you double th emolarity of acetone in the reaction mixture, keeping the total volume at 50mL and keeping the same concentrations of H+ ion and I2 as in the original mixture?

Answer 1

5mL 4.0 M acteon + 10ml 1.0 M HCl + 10 mL 0.0050 M I2 + 25 mL H2O
no of moles of acetone A = molarity * volume in L
                         = 4*0.005
                         = 0.02 moles of acetone
b. total volume of solution is 50ml = 0.05L
molarity of acetone in the reaction mixture
molarity = no of moles/volume
         = 0.02/0.05 = 0.4M
c. initialymolarity of acetone taken the 8M
no of moles of acetone is double and molarity of acetone in the reaction mixture double