Assignment # 12 Chemistry 0500A Key Fall 2011 • Chapter 14: 32, 38, 42, 44, 48, 50, 54b & d, 58a & c, 60, 66, 70 and 74Due Monday Dec 5 32.
(a)~35 °C
38.
(a)supersaturated at 0 °C
(b)saturated at 25 °C
(c)unsaturated at 100 °C
42.
(a)~100 g solute remains in solution
(b)~20 g solute (120 g – 100 g) crystallizes from solution
44.
(b)
~43 °C
mass of solute
mass of solution × 100% = m/m %
1.25 g NaCl
(a) 100.0 g solution × 100% = 1.25%
2.50 g K2Cr2O7
(b) 95.0 g solution
× 100% = 2.63%
10.0 g CaCl2
(c) 250.0 g solution × 100% = 4.00%
48.
65.0 g sugar
(d) 125.0 g solution
× 100% = 52.0%
(a) 35.0 g H2SO4
×
100 g solution
5.00 g H2SO4
= 7.00 × 102 g solution
100 g solution
(b) 10.5 g HC2H3O2 × 4.50 g HC H O = 233 g solution 2 3 2
50.
(a) 10.0 g solution
(b) 50.0 g solution
×
×
2.50 g K2CO3
100 g solution
5.00 g Li2SO4
100 g solution
= 0.250 g K2CO3
= 2.50 g Li2SO4
54. (b) MM of Na2CrO4 = 2(22.99 g/mol) + 52.00 g/mol + 4(16.00 g/mol)
= 161.98 g/mol
75.0 mL solution = 0.0750 L solution
1.00 g Na2CrO4
1 mol Na2CrO4
×
0.0750 L solution
161.98 g Na2CrO4
= 0.0823 M Na2CrO4
(d) MM of Li2CO3 = 2(6.94 g/mol) + 12.01 g/mol + 3(16.00 g/mol)
= 73.89 g/mol
250.0 mL solution = 0.2500 L solution
20.0 g Li2CO3
1 mol Li2CO3
×
0.2500 L solution
73.89 g Li2CO3
= 1.08 M Li2CO3
58. (a) MM of KNO3 = 39.10 g/mol + 14.01 g/mol + 3(16.00 g/mol)
= 101.11 g/mol
1 mol KNO3
1 L solution
2.50 g KNO3 × 101.11 g KNO × 0.325 mol KNO = 0.0761 L solution
3
3
(b)MM of AlBr3 = 26.98 g/mol
+ 3(79.90 g/mol)
= 266.68 g/mol
1 mol AlBr3
1 L solution
2.50 g AlBr3 × 266.68 g AlBr × 0.325 mol AlBr = 0.0288 L solution
3
3
(c) MM of Co(C2H3O2)2 = 58.93 g/mol + 4(12.01 g/mol) + 6(1.01 g/mol)
+ 4(16.00 g/mol) = 177.03 g/mol
1 mol Co(C2H3O2)2
1 L solution
2.50 g Co(C2H3O2)2 × 177.03 g Co(C H O ) × 1.00 mol Co(C H O )
2 3 2 2
2 3 2 2
60.
= 0.0141 L solution
(a) MM of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
1.00 L solution ×
0.100 mol NaOH
40.00 g NaOH
×
1 L solution
1 mol NaOH = 4.00 g NaOH
(b) MM of LiHCO3 = 6.94 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol) =
67.96 g/mol
1.00 L solution ×
0.100 mol LiHCO3
67.96 g LiHCO3
×
1 L solution
1 mol LiHCO3 = 6.80 g LiHCO3
(c) MM of CuCl2 = 63.55 g/mol + 2(35.45 g/mol) = 134.45 g/mol
0.500 mol CuCl2
134.45 g CuCl2
25.0 mL solution × 1000 mL solution × 1 mol CuCl
= 1.68 g CuCl2
2
(d) MM of KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)
= 158.04 g/mol
158.04 g KMnO4
0.500 mol KMnO4
25.0 mL solution × 1000 mL solution × 1 mol KMnO
= 1.98 g KMnO4
4
66.
50.0 mL solution
×
17 moL HC2H3O2
1000 mL solution =
=
M1
×
V1 =
17 M ∞ 50.0 mL
500.0 mL
74.
M2
=
×
0.85 mol HC2H3O2
1.7 M HC2H3O2
V2
1.7 M HC2H3O2
0.63 g Cl2
500.0 mL solution × 100 mL solution = 3.2 g Cl2