AP CHEM HW: p. 532 / 29, 30, 33, 35, 37
29)
Given: 10.0 g H3PO4, 100.0 mL H2O, 104 mL soln
10.0 g H3PO4 = 0.102 mol H3PO4
100.0 g H2O = 5.549 mol H2O
30)
mass soln = 110.0 g
V soln = 104 mL
Given: 40.0 % C2H6O2, dsoln = 1.05 g/mL
- assume 100 g soln
V soln 
40.0 g C2H6O2 = 0.644 mol C2H6O2
60.0 g H2O
= 3.33 mol H2O
33)
mass soln = 100 g
V soln = 95.2 mL
Given: 25 mL C5H12 (d = 0.63 g/mL), 45 mL C6H14 (d = 0.66 g/mL), volumes additive
mass C5H12 
(
)(
)
mass C6H14 
(
)(
)
16 g C5H12 = 0.22 mol C5H12
30. g C6H14 = 0.35 mol C6H14
(
)
mass soln = 46 g
V soln = 70. mL
(
)
35)
Given: 12.5% C2H5OH (shorthand: EtOH) by volume, dEtOH = 0.789 g/mL
- assume 100 mL soln  12.5 mL EtOH, 87.5 mL H2O
9.86 g EtOH = 0.214 mol EtOH
87.5 g H2O = 4.86 mol H2O
(
37)
)(
(
mass EtOH 
)
)
mass soln = 97.4 g
V soln = 100 mL
(
)
Given: 1.37 M H3C6H5O7, dsoln = 1.10 g/mL
- assume 1 L soln
)(
(
mass soln 
mass soln = ̅
V soln = 1000 mL
1.37 mol H3C6H5O7 = 263 g H3C6H5O7
47 mol H2O
= 840 g H2O
(
)
(
)
̅
N = nM = (3)(1.37 M) = 4.11 N H3C6H5O7
)
̅