Chapter 13 Homework 13.42 Calculate the

Chapter 13 Homework
13.42 Calculate the concentration of each solution in mass percent.
a. 132 g KCl in 598 g H2O
SOLUTION: % 𝑏𝑦 π‘šπ‘Žπ‘ π‘  =
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘‘π‘œπ‘‘π‘Žπ‘™ π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
% 𝑏𝑦 π‘šπ‘Žπ‘ π‘  =
× 100
132 𝑔 𝐾𝐢𝑙
× 100 = 18.1% 𝑏𝑦 π‘šπ‘Žπ‘ π‘ 
132 𝑔 𝐾𝐢𝑙 + 598 𝑔 𝐻2 𝑂
b. 22.3 mg KNO3 in 2.84 g H2O
22.3 π‘šπ‘”
% 𝑏𝑦 π‘šπ‘Žπ‘ π‘  =
1𝑔
= 0.0223 𝑔
1000 π‘šπ‘”
0.0223 𝑔 𝐾𝑁𝑂3
× 100 = 0.779 % 𝑏𝑦 π‘šπ‘Žπ‘ π‘ 
0.0223 𝑔 𝐾𝑁𝑂3 + 2.84 𝑔 𝐻2 𝑂
c. 8.72 g C2H6O in 76.1 g H2O
% 𝑏𝑦 π‘šπ‘Žπ‘ π‘  =
8.72 𝑔 𝐢2 𝐻6 𝑂
× 100 = 10.3% 𝑏𝑦 π‘šπ‘Žπ‘ π‘ 
8.72 𝑔 𝐢2 𝐻6 𝑂 + 76.1 𝑔 𝐻2 𝑂
13.50 Determine the amount of potassium chloride in each solution.
a. 19.7 g of a solution containing 1.08% KCl by mass
SOLUTION: UNITS! UNITS! UNITS!
Remember % 𝑏𝑦 π‘šπ‘Žπ‘ π‘  ≑
𝑔 π‘ π‘œπ‘™π‘’π‘‘π‘’
100 𝑔 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
19.7 𝑔 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
1.08 𝑔 𝐾𝐢𝑙
= 0.213 𝑔 𝐾𝐢𝑙
100 𝑔 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
b. 23.2 kg of a solution containing 18.7% KCl by mass
23.2 π‘˜π‘” π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
1000 𝑔 18.7 𝑔 𝐾𝐢𝑙
= 4338 𝑔 𝐾𝐢𝑙
1 π‘˜π‘” 100 𝑔 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
OR
23.2 π‘˜π‘” π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
18.7 π‘˜π‘” 𝐾𝐢𝑙
= 4.338 π‘˜π‘” 𝐾𝐢𝑙
100 π‘˜π‘” π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
c. 38 mg or a solution containing 12% KCl by mass
38 π‘šπ‘” π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
12 π‘šπ‘” 𝐾𝐢𝑙
= 4.56 π‘šπ‘” 𝐾𝐢𝑙
100 π‘šπ‘” π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
13.60 Calculate the molarity of each solution.
a. 1.54 mol of LiCl in 22.2 L of solution
1.54 π‘šπ‘œπ‘™ 𝐿𝑖𝐢𝑙
= 0.069 𝑀 𝐿𝑖𝐢𝑙
22.2 𝐿 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
b. 0.101 mol of LiNO3 in 6.4 L of solution
0.101 π‘šπ‘œπ‘™ 𝐿𝑖𝑁𝑂3
= 0.0158 𝑀 𝐿𝑖𝑁𝑂3
6.4 𝐿
d. 0.0323 mol of glucose in 76.2 mL of solution
0.0323 π‘šπ‘œπ‘™ π‘”π‘™π‘’π‘π‘œπ‘ π‘’
= 0.424 𝑀 π‘”π‘™π‘’π‘π‘œπ‘ π‘’
1𝐿
76.2 π‘šπΏ
1000 π‘šπΏ
13.62 Calculate the molarity of each solution.
a. 33.2 g of KCl in 0.895 L of solution
1 π‘šπ‘œπ‘™ 𝐾𝐢𝑙
74.55 𝑔
= 0.498 𝑀 𝐾𝐢𝑙
0.895 𝐿
33.2 𝑔 𝐾𝐢𝑙
b. 61.3 g of C2H6O in 3.4 L of solution
1 π‘šπ‘œπ‘™ 𝐢2 𝐻6 𝑂
46.068 𝑔
= 0.391 𝑀 𝐢2 𝐻6 𝑂
3.4 𝐿
61.3 𝑔 𝐢2 𝐻6 𝑂
c. 38.2 mg of KI in 112 mL of solution
1 𝑔 1 π‘šπ‘œπ‘™πΎπΌ
38.2 π‘šπ‘” 𝐾𝐼 1000 π‘šπ‘” 166.1 𝑔
= 0.00152 𝑀 𝐾𝐼
1𝐿
112 π‘šπΏ 1000 π‘šπΏ
13.82 A 3.5 L sample of a 5.8 M NaCl solution is diluted to 55 L. What is the molarity of the diluted
solution?
SOLUTION:
3.5 𝐿 π‘π‘ŽπΆπ‘™ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
5.8 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
= 20.3 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
𝐿
20.3 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
= 0.369 𝑀 π‘π‘ŽπΆπ‘™
55 𝐿 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
13.92 Consider the reaction:
K2S (aq) + Co(NO3)2 (aq) β†’ 2 KNO3 (aq) + CoS (s)
What volume of 0.225 M K2S solution is required to completely react with 175 mL of 0.115 M Co(NO3)2?
1 𝐿 0.115 π‘šπ‘œπ‘™ πΆπ‘œ(𝑁𝑂3 )2
1 π‘šπ‘œπ‘™ 𝐾2 𝑆
1𝐿
1000 π‘šπΏ
𝐿
1 π‘šπ‘œπ‘™ πΆπ‘œ(𝑁𝑂3 )2 0.225 π‘šπ‘œπ‘™ 𝐾2 𝑆
= 0.089 𝐿 𝐾2 𝑆 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
175 π‘šπΏ πΆπ‘œ(𝑁𝑂3 )2