answers_even_chapter14

answers_even_chapter14 - Chapter 14 14.56 Solutions and...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/09/2012 for the course CHEMISTRY 101 taught by Professor Endy during the Spring '11 term at Harvard.

Page1 / 2

answers_even_chapter14 - Chapter 14 14.56 Solutions and...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online