Question

Answer all the questions: Part B(40 points) 1. What volume (in milliliters) of a 0.315 M NaOH solution contains 622 g of NaOH? 2. Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 ml? 3. How would you prepare 2.00 x 10 ml of a 0.866 M NaOH solution, starting with a 5.07 M stock solution?

Answer 1

1)

moles of NaOH = 6.22 / 40 = 0.156 mol

moles = concentration x volume

0.315 x V = 0.156

V = 0.496

volume of NaOH = 494 mL

2)

concentration = moles / volume

                      = 0.715 / 0.495

Molarity = 1.44 M

3)

volume V2 = 2.00 x 10^2 mL

molarity C2 = 0.866 M

concentration C1 = 5.07 M

C1 V1 = C2 V2

5.07 x V1 = 0.866 x 200

V1 = 34.16 ml

volume of stock solution = 34.2 mL

by adding 165.8 mL of water to stock solution of 34.2 mL of NaOH to prepare 0.866 M NaOH