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# answers_even_chapter14 - Chapter 14 14.56 Solutions and...

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Chapter 14 Solutions and Their Behavior 245 14.56 10.7 mol NaOH á 40.00 g 1 mol NaOH = 428 g NaOH 1 × 10 3 g H 2 O á 1 mol H 2 O 18.02 g = 55.5 mol H 2 O (428 g + 1000 g) á 1 cm 3 1.33 g á 1 mL 1 cm 3 á 1 L 10 3 mL = 1.07 L solution (a) X NaOH = 10.7 mol NaOH 10.7 mol NaOH + 55.5 mol H 2 O = 0.162 (b) Weight % = 428 g NaOH 428 g NaOH + 1000 g H 2 O 100% = 30.0% (c) M NaOH = 10.7 mol NaOH 1.07 L solution = 9.97 M 14.70 5 g á 1 mol CH 3 CO 2 H 60.0 g = 0.08 mol CH 3 CO 2 H 95 g á 1 mol H 2 O 18.0 g = 5.3 mol H 2 O X CH 3 CO 2 H = 0.08 mol CH 3 CO 2 H 0.08 mol CH 3 CO 2 H + 5.8 mol H 2 O = 0.02 m CH 3 CO 2 H = 0.08 mol CH 3 CO 2 H 0.095kg H 2 O = 0.9 m 5 × 10 Ğ 3 mg CH 3 CO 2 H 100 g solution á 1 mL 1 g á 1 L 10 3 mL = 0.05 mg/L = 0.05 ppm Calculating molarity requires knowing the total volume of the solution. Without knowing the density of the acetic acid solution it is impossible to calculate the molarity of the solution.

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answers_even_chapter14 - Chapter 14 14.56 Solutions and...

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