1 (10)

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10 Solutions Manual for Analytical Chemistry 2.1
6. Using the atomic weights from Appendix 18, we ind that the formula 
weight for Ni(C4H14N4O4)2 is
( . ) ( . ) ( . )
( . ) ( . ) .
58 693 8 12 011 14 1 008
14 007 15 999 288 914 4 7 g/mole
# #
# #
+ + +
+ =
7. First we convert the mass of Cl– to moles of Cl–
.
.
256
1000
1
35 45
1 7 221 10
mg Cl
mg Cl
g Cl
g Cl
mol Cl mol Cl3
# #
#=
-
-
-
-
-
- -
 and then the moles of Cl– to the moles of BaCl2
. .7 221 10
2
1 3 610 10mol Cl
mol Cl
mol BaCl mol BaCl3 32
2# # #=
- -
-
-
 and inally the moles of BaCl2 to the volume of our BaCl2 solution
.
.
.
3 610 10
0 217
1
1
1000 16 6
mol BaCl
mol BaCl
L
L
mL mL
3
2
2
# #
# =
-
8. We can express a part per million in several ways—this is why some 
organizations recommend against using the abbreviation ppm—but 
here we must assume that the density of the solution is 1.00 g/mL and 
that ppm means mg/L or µg/mL. As molarity is expressed as mol/L, 
we will use mg/L as our starting point; thus
L
0.28 mg Pb
1000 mg
1 g
207.2 g Pb
1 mol Pb 1.4 10 M Pb6
# # #=
-
9. (a) he molarity of 37.0% w/w HCl is
1.00 10 g solution
37.0 g HCl
ml solution
1.18 g solution
L
1000 mL
36.46 g HCl
1 mol HCl 12.0 M HCl
2
#
# #
# =
 (b) To calculate the mass and volume of solution we begin with the 
molarity calculated in part (a). To avoid any errors due to rounding 
the molarity down to three signiicant, we will return one additional 
signiicant igure, taking the molarity as 11.97 M.
0.315 mol HCl
11.97 mol HCl
1 L
L
1000 mL
mL
1.18 g solution
31.1 g
# #
# =
.26 30.315 mol HCl
11.97 mol HCl
1 L
L
1000 mL mL# # =
For problems in this chapter, all formu-
la weights are reported to the number of 
signiicant igures allowed by the atomic 
weights in Appendix 18. As a compound’s 
formula weight rarely limits the uncer-
tainty in a calculation, in later chapters 
usually we will round formula weights to 
a smaller number of signiicant igures, 
chosen such that it does not limit the cal-
culation’s uncertainty.
For problem 7 we include an extra signii-
cant igure in each of the calculation’s irst 
two steps to avoid the possibility of intro-
ducing a small error in the inal calcula-
tion as a result of rounding. If we need to 
report the result for an intermediate calcu-
lation, then we round that result appropri-
ately; thus, we need to isolate 3.61×10
–3
 
mol of BaCl2.