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

Question

In a reaction involving the iodination of acetone, the following volumes were used 10 make up the reaction mixture: 5 mL 4.0 M acetone + 10 mL 1.0 M HCI + 10 mL 0.0050 M I_2 + 25 mL H_2O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = M_A times V, where M_A 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, M_A = moles of 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 I_2 as in the original mixture?

Explanation / Answer

(a)

Given that,

Molarity of acetone = 4.0 M

Volume of acetone solution used = 5 mL = 0.005 L

Moles of acetone = Molarity of acetone * volume of acetone solution in L

Moles of acetone = 4.0 * 0.005 = 0.020 mol

(b)

Number of moles of acetone in the reaction mixture = 0.020 mol

Volume of reaction mixture = 50 mL = 0.050 L

Therefore,

Molarity of acetone in the reaction mixture = number of moles of acetone / volume of acetone

Molarity of acetone = 0.020 / 0.050 = 0.40 M

(c)

Use double amount of acetone i.e 10 mL of acetone but 20 mL of water to keep total volume of reaction mixture as 50 mL

10 mL of 4.0 M acetone + 10 mL of 1.0 M HCl + 10 mL of 0.0050 M I2 + 20 mL H2O

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