Student Solution Manual
Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
Some of the answers below may differ in format but have the same value as your result. Please
check with your instructor if a specific format is desired.
Chapter 2
2-1. Define
Answers:
(a) molar mass.
The molar mass is the mass in grams of one mole of a chemical species.
(c) millimolar mass.
The millimolar mass is the mass in grams of one millimole of a chemical species.
2-3. Give two examples of units derived from the fundamental base SI units.
Solution:
3
1000 mL 1 cm3 m
The liter: 1 L 10 m
3 3
1L mL 100 cm
1 mol L 1 mol
Molar concentration: 1 M
L 103 m3 103 m3
2-4. Simplify the following quantities using a unit with an appropriate prefix:
Solutions:
(a) 5.8 108 Hz.
MHz
5.8 108 Hz 580 MHz
106 Hz
(c) 9.31 107 mol.
mol
9.31 107 mol 93.1 mol
106 mol
(e) 396 106 nm.
mm
3.96 106 nm 3.96 mm
106 nm
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part. 1
, Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
2-5. Why is 1 g no longer exactly 1 mole of unified atomic mass units?
Answer:
The dalton is defined as 1/12 the mass of a neutral 12C atom. With the redefinition of SI
base units in 2019, the definition of the dalton remained the same. However, the
definition of the mole and the kilogram changed in such a way that the molar mass unit
is no longer exactly 1 g/mol.
2-7. Find the number of Na+ ions in 2.75 g of Na3PO4?
Solution:
2.75 g Na PO 1 mol Na3PO4 3 mol Na 6.022 10 Na 3.03 1022 Na
23
3 4
163.94 g mol Na3PO4 mol Na
2-9. Find the amount of the indicated element (in moles) in
Solutions:
(a) 5.32 g of B2O3.
5.32 g B O 2 mol B mol B2O3 0.153 mol B
2 3
mol B2O3 69.62 g B2O3
(b) 195.7 mg of Na2B4O7 10H2O.
195.7 mg Na B O 10H O g 7 mol O
2 4 7 2
1000 mg mol Na2B4O7 10H2O
mol Na2B4O 7 10H2O
3.59 103 mol O = 3.59 mmol
381.37 g
(c) 4.96 g of Mn3O4.
mol Mn3O4
4 3 mol Mn 6.50 102 mol Mn
4.96 g Mn3O4 228.81 g Mn3O
mol Mn3O 4
(d) 333 mg of CaC2O 4.
2 mol C
g 5.20 103 mol C
333 mg CaC O
2 4
mol CaC2O4
1000 mg 128.10 g CaC2O4 mol CaC2O4
5.20 mmol
2-11. Find the number of millimoles of solute in
Solutions:
(a) 2.00 L of 0.0449 MKMnO .
4
0.0449 mol KMnO4 1000 mmol
2.00 L 89.8 mmol KMnO 4
L mol
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part. 2
, Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
(b) 750 mL of 535 1023 MKSCN.
5.35 103 M KSCN 1000 mmol L
750 mL 4.01 mmol KSCN
L mol 1000 mL
(c) 3.50 L of a solution that contains 6.23 ppm of CuSO 4.
6.23 mg CuSO4 g mol CuSO4 1000 mmol
3.50 L 0.137 mmol CuSO
L 1000 mg 159.61 g CuSO4 mol 4
(d) 250 mL of 0.414 mM KCl.
0.414 mmol KCl 1L
250 mL 0.104 mmol KCl
L 1000 mL
2-13. What is the mass in milligrams of
Solutions:
(a) 0.367 mol of HNO3?
63.01 g HNO3 1000 mg
0.367 mol HNO = 2.31 104 mg HNO
3 3
mol HNO3 g
(b) 245 mmol of MgO?
mol 40.30 g MgO 1000 mg
245 mm ol MgO 9.87 103 mg MgO
1000 mm ol mol MgO g
(c) 12.5 mol of NH4NO3 ?
80.04 g NH4 NO 3 1000 mg
12.5 mol NH NO 1.00 106 mg NH NO
3
4 3 mol NH4 NO3 g 4
(d) 4.95 mol of NH4 2 eNO3 6 548.23 g/mol?
548.23 g (NH4 )2Ce(NO3 )6 1000 mg
4.95 mol (NH4 )2Ce(NO3 )6
mol (NH4 )2Ce(NO3 )6 g
2.71 106 mg (NH4 )2Ce(NO3 )6
2-15. What is the mass in milligrams of solute in
Solutions:
(a) 16.0 mL of 0.350 M sucrose (342 g/mol)?
0.350 mol sucrose L 342 g sucrose 1000 mg
L 1000 mL mol sucrose g
16.0 mL 1.92 103 mg sucrose
© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part. 3
, Student Solution Manual: Skoog et al., Fundamentals of Analytical Chemistry, 10e,
© 2022, 978-0-357-45055-0, Chapter 2: Calculations Used in Analytical Chemistry
(b) 1.92 L of 376 1023 M H 2O 2 ?
3.76 10 3 mol H2O2 34.02 g H2O2 1000 mg
1.92 L 246 mg H O
L mol H2 O2 g 2 2
2-16. What is the mass in grams of solute in
Solutions:
(a) 250 mL of 0.264 M H2O2 ?
0.264 mol H2O2 L 34.02 g H2O 2
250 mL 2.25 g H 2O 2
L 1000 mL mol H2O2
(b) 37.0 mL of 5.75 104 M benzoic acid (122 g/mol)?
5.75 104 mol benzoicacid L 122 g benzoicacid
L 1000 mL mol benzoicacid
37.0 mL 2.60 103 g benzoicacid
2-17. Calculate the p-value for each of the listed ions in the following:
Solutions:
(a) Na1 , Cl , and OH in a solution that is 0.0635 M in NaCl and 0.0403 M in NaOH.
pNa log(0.0635 0.0403) log(0.1038) 0.9838
pCl log(0.0635) 1.197
pOH log (0.0403) 1.395
(c) H, Cl , and Zn21 in a solution that is 0.400 M in HCl and 0.100 M in ZnCl 2.
pH log(0.400) 0.398
pCl log(0.400 2 0.100) log(0.600) 0.222
pZn log(0.100) 1.00
(e) K, OH, and Fe CN in a solution that is 1.62 107 M in K FeCN and 5.12 107 M
42
6 4 6
in KOH.
pK log(4 1.62 107 5.12 107 ) log(1.16 106 ) 5.94
pOH log(5.12 107 ) = 6.291
pFe(CN)6 log(1.62 107 ) = 6.790
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