Question

3. You dissolve the sample of FD&C Red 40 in enough water to make 250.0 mL of solution. We call this the stock solution What is the concentration of the stock solution in terms of molarity? 8.7x10-5 M You are correct. Your receipt no. is 156-2105 4. You make two dilutions. You take 87.60 mL of the stock solution and dilute to 500.0 mL to make Solution #1. You then add 65.28 mL of Solution 1 to 37.34 mL of water to make Solution #2. What is the concentration of the Solution #1 in terms of molarity? 1.525x10-5 M You are correct Your receipt no. is 156-19076) PreviousTies 5, What is the concentration of the Solution #2 in terms of molarity? 9.699x10-6 M You are correct. Your receipt no. is 156-4385 6. You measure the absorbance of each solution (atx 524 nm) and obtain the following results: Solution Absorbance Stock solution Solution #1 Solution #2 3.23 0.560 0.355 Determine the average epsilon for FD&C Red 40 709000000 M-cm Subrnit Answer Incorrect. Tries 12/99 Previous Tries

Answer 1

A = ECl. Stock solution: = E times 8.7 times 10^-5 E_1 = 37126.44 M^-1 cm^-1 solution 1: - 0.560 = E times 1.525 times 10^-5 E_2 = 36721.31 M^-1 cm^-1 solution 2: - 0.355 = E times 9.699 times 10^-6 E_3 = 36601.71 M^- cm^-1 Average E = E_1 + E_2 + E_3/3 = 3.682 times 10^4 M^-1 cm^-1