3 4 4 4 4 4 4 mol agclo 29993 g agclo 1 mol agclo 10

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3 4 4 4 4 4 4 ? mol AgClO 29.993 g AgClO 1 mol AgClO 10 mL Molarity = = 1 L AgClO soln 50.0 mL AgClO soln 207.3185 g AgClO 1 L = 4 4 2.893 mol AgClO 1 L AgClO soln = 2.893 M AgClO 4 Exercise 5.11 – General Stoichiometry: What is the maximum number of grams of silver chloride that will precipitate from a solution made by mixing 25.00 mL of 0.050 M MgCl 2 with an excess a AgNO 3 solution? 2 2 3 2 2 0.050 mol MgCl 2 mol AgCl 143.3209 g AgCl ? g AgCl = 25.0 mL MgCl 1 mol MgCl 1 mol AgCl 10 mL MgCl = 0.36 g AgCl Exercise 5.12 – Titration Problem: When 34.2 mL of a 1.02 M NaOH solution is added from a burette to 25.00 mL of a phosphoric acid solution that contains phenolphthalein, the solution changes from colorless to red. a. What is the titrant for this process? NaOH is the titrant. b. What is the molarity of the phosphoric acid? 3 3 4 3 4 3 3 4 3 4 ? mol H PO 1 mol H PO 34.2 mL NaOH soln 1.02 mol NaOH 10 mL = L H PO soln 25.00 mL H PO soln 3 mol NaOH 1 L 10 mL NaOH soln = 0.465 M H 3 PO 4 Exercise 5.13 – Making solution from Solid: An experiment calls for a total of 1.50 L of 0.200 M KMnO 4 for a class of chemistry students. How would this solution be made from pure, solid potassium permanganate and water? 4 4 4 4 0.200 mol KMnO 158.0339 g KMnO ? g KMnO = 1.50 L soln 1 L soln 1 mol KMnO = 47.4 g KMnO 4 Dissolve 47.4 g KMnO 4 in a minimum amount of water and dilute with water to 1.50 L total. Copyright 2004 Mark Bishop
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Chapter 5 Exercise Key Exercise 5.14 – Dilution Problems: What is the molarity of a solution made by diluting 5.00 mL of a 14.8 M NH 3 solution to 75.0 mL? 3 3 3 3 ? mol NH 14.8 mol NH 5.00 mL conc. soln 10 mL = L dil. soln 75.00 mL dil. soln 1 L 10 mL conc. soln or ( ) C C C C D D D D 14.8 M 5.00 mL M V M V = M V M = = V 75.0 mL = 0.987 M NH 3 Exercise 5.15 – Making solution from Concentrated Acid: How would you make 250.0 milliliters of 2.00 M acetic acid from concentrated acetic acid (called glacial acetic acid) that is 17.4 M HC 2 H 3 O 2 ? 3 2 3 2 3 2 3 2 2.00 mol HC H O 10 mL conc. soln ? mL conc. soln = 250.0 mL dil. soln 17.4 mol HC H O 10 mL dil. soln or ( ) D D C C D D C C 2.00 M 250.0 mL M V M V = M V V = = M 17.4 M = 28.7 mL conc. soln Carefully add 28.7 mL of 17.4 M HC 2 H 3 O 2 to about 200 mL of water, mix, and dilute with water to 250.0 mL total.
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