Question

Your instructor will supply the following solutions (for each pair): Solution A50 mL of 1.0 x 103 Fe2 in 0.1 M HC1 in a dry 100 mL beaker Solution B 60 mL of 1.0 x 10-3 M o-phen in a dry 50 mL beaker Solution C 10 mL of freshly prepared aqueous 1 percent hydroquinone solution ina dry 50 mL beaker 10 mL of 1 M HCI solution in a dry 50 mL beaker Solution D

Determine the number of millimoles of Fe^2+ and o-phen in each solution and obtain the total millimoles.

Answer 1

Solution 1:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 0

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 10 * 1.0 * 10^-3 mmoles = 0.01 mmoles

Solution 2:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 1 * 1.0 * 10^-3 mmole = 0.001 mmol

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 9 * 1.0 * 10^-3 mmoles = 0.009 mmoles

Solution 3:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 2 * 1.0 * 10^-3 mmole = 0.002 mmol

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 8 * 1.0 * 10^-3 mmoles = 0.008 mmoles

Solution 4:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 4 * 1.0 * 10^-3 mmole = 0.004 mmol

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 6 * 1.0 * 10^-3 mmoles = 0.006 mmoles

Solution 5:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 6 * 1.0 * 10^-3 mmole = 0.006 mmol

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 4 * 1.0 * 10^-3 mmoles = 0.004 mmoles

Solution 6:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 8 * 1.0 * 10^-3 mmole = 0.008 mmol

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 2 * 1.0 * 10^-3 mmoles = 0.002 mmoles

Solution 7:

mmoles of Fe⁺² = volume of solution in ml * Molarity of the solution = 10 * 1.0 * 10^-3 mmole = 0.01 mmol

mmoles of 0- phen = volume of solution in ml * Molarity of the solution = 0