Chapter 2
Solutions
21
1.
How many grams of CuSO
4
are required to make 650. mL of a 0.115M solution?
4
4
4
4
1 L solution
0.115 mol CuSO
159.608 g CuSO
650. mL solution
11.9 g CuSO
1000 mL solution
L solution
1 mol CuSO
×
×
×
=
3.
How many grams of Na
2
SO
4
are required to make 90.0 mL of a solution that is 0.200 M in Na
1+
?
1
2
4
2
4
2
4
1
2
4
1 L solution
0.200 mol Na
1 mol Na
SO
142.042 g Na
SO
90.0 mL solution
1.28 g Na
SO
1000 mL solution
L solution
1 mol Na
SO
2 mol Na
+
+
×
×
×
×
=
5.
How many mmoles of chloride ion are in 55.0 mL of 0.0688 M BaCl
2
solution?
1
1
2
2
0.0688 mmol BaCl
2 mmol Cl
55.0 mL solution
7.57 mmol Cl
1 mL solution
1 mmol BaCl
×
×
=
7.
How many mL of 0.124 M
Ba(OH)
2
are required to deliver 38.6 mmol of hydroxide ion ?
1
2
1
2
1 mmol Ba(OH)
1 mL solution
38.6 mmol OH
156 mL solution
0.124 mmol Ba(OH)
2 mmol OH
×
×
=
9.
How many grams of sodium should be added to 15.0 g Hg to make a mixture in which the mole fraction of sodium
is 0.800?
Na
1 mol Hg
moles of mercury =15.0 g Hg
0.0748 mol Hg
200. 59 g Hg
n
Use moles of Hg and the mole fraction equation: X
0.800
,
0.0748
n
0.0598
rearrange to eliminate denominator: 0.0598
0.800n
n
n
0
×
=
=
=
+
+
=
⇒
=
= 0.299 mol Na
.200
22.99 g Na
Calculate the grams of Na:
0.299 mol Na
6.88 g Na
1 mol Na
×
=
11. The density of a 1.140M solution of NH
4
Cl at 20
o
C is 1.0186 g/mL.
What mass of water does 100. mL of this
solution contain?
The total mass of this solution is
1.0186 g/mL x 100. mL = 102 (three significant figures)
The mass of NH
4
Cl in this solution is
4
4
4
4
1.140 mol NH
Cl
53.49 g
0.100 L
= 6.10g
NH Cl
1 L NH
Cl
1 mol NH
Cl
×
×
The mass of water is then the difference between the total mass of the solution and the mass of the NH
4
Cl:
102 g solution  6 g NH
4
Cl = 96 g water.
13. What mass of Fe(ClO
4
)
3
is required to make 275 mL of a solution that is 0.100 M in ClO
4
1
?
1
4 3
4 3
4
4 3
1
4
1 mol Fe(ClO
)
354 g Fe(ClO
)
0.100 mol ClO
0.0275 L
3.25 g Fe(ClO
)
1 L
1 mol
3 mol ClO
−
×
×
×
=
15. How many grams of CaCl
2
should be added to 50.0 g of water to make a solution in which the mole fraction of Cl
1
is 0.150?
First get the number of moles of water:
2
2
2
2
1 mol H O
50.0 g H O
= 2.78 mol H O
18.0 g H O
×
If n mol CaCl
2
are dissolved, n mol Ca
2+
and 2n mol Cl
1
form in solution, or 3n mol ions.
1
2
2
2
2
The mole fraction of chloride ion is 0.150, so
2n mol Cl
0.150
3n mol ions
2.78 mol of H O
Rearrange and solve for n:
0.450n
0.417
2n
n = 0.269 mol CaCl
111 g CaCl
0.269 mol CaCl
1 mol
=
+
+
=
⇒
×
2
29.9 g CaCl
=
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Solutions
22
17. A concentrated solution of phosphoric acid is 75% H
3
PO
4
by mass and has a density of 1.57 g/mL.
a)
What is the molarity of the concentrated phosphoric acid?
Assume 100. g of concentrated phosphoric acid then 75% translates to 75 g of H
3
PO
4
.
Calculate the moles of
H
3
PO
4
.
3
4
3
4
3
4
1 mol H PO
75 g H PO
=0.77 mol H PO
97.994 g
×
Use the density to calculate the volume of 100 g of concentrated phosphoric acid.
3
4
3
4
1 mL
100. g conc H PO
=63.7 mL conc H PO
1.57 g
×
The molarity of the solution can then be calculated,
3
4
3
4
0.77 mol H PO
=12. M H PO
0.0637 L solution
b)
How many mL of the concentrated acid would be required to prepare 1.5 L of a 0.20M solution?
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 Spring '10
 Wilson
 Chemistry, Mole, mol Na, ml, mol Si

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