Chapter 4 solutions - Fundamentals of Analytical Chemistry 8th ed Chapter 4 Chapter 4 4-1(a The millimole is an amount of a chemical species such as an

Chapter 4 solutions - Fundamentals of Analytical Chemistry...

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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 Chapter 4 4-1 (a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule or an electron. A millimole contains millimole particles 10 02 . 6 millimole mole 10 mole particles 10 02 . 6 20 3 23 × = * × - (b) The molar mass is the mass in grams of one mole of a chemical species. (c) The millimolar mass is the mass in grams of one millimole of a chemical species. (d) Parts per million, c ppm , is a term expressing the concentration of very dilute solutions. Thus, c ppm ppm 10 solution of mass solute of mass 6 × = The units of mass in the numerator and the denominator must be the same. 4-2 The species molarity of a solution expresses the equilibrium concentration of a chemical species in terms of moles per liter. The analytical molarity of a solution gives the total number of moles of a solute in one liter. The species molarity takes into account chemical reactions that occur in solution. The analytical molarity specifies how the solution was prepared, but does not account for any subsequent reactions. 4-3 3 3 3 3 m 10 cm 100 m mL cm 1 L 1 mL 1000 1L - = × × = 3 3 3 3 m 10 mole 1 m 10 L L mole 1 M 1 - - = × =
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 4-4 (a) kHz 320 Hz 1000 kHz Hz 10 2 . 3 5 = × × (b) ng 6 . 45 g ng 10 g 10 56 . 4 9 8 = × × - (c) mmol 843 mol 10 mmol mol 10 43 . 8 3 5 = μ × μ × (d) Ms 5 . 6 s 10 Ms s 10 5 . 6 6 6 = × × (e) m 6 . 89 nm 10 m nm 10 96 . 8 3 4 μ = μ × × (f) kg 72 g 1000 kg g 000 , 72 = × 4-5 + + + + × = × × × × Na 10 98 . 5 Na mol Na 10 02 . 6 PO Na mol Na mol 3 g 94 . 163 PO Na mol 1 PO Na g 43 . 5 22 23 4 3 4 3 4 3 4-6 + + + + × = × × × K 10 22 . 1 K mol K 10 02 . 6 PO K mol K mol 3 PO K mol 76 . 6 25 23 4 3 4 3 4-7 (a) 3 2 3 2 3 2 3 2 O B mol 0712 . 0 O B g 62 . 69 O B mol O B g 96 . 4 = × (b) O 10H O B Na mol 10 73 . 8 381.37g O 10H O B Na mol mg 1000 g O H 10 O B Na mg 333 2 7 4 2 4 2 7 4 2 2 7 4 2 × = × × - (c) 4 3 4 3 4 3 4 3 O Mn mol 0382 . 0 O Mn g 81 . 228 O Mn mol O Mn g 75 . 8 = × (d) 4 2 3 4 2 4 2 4 2 O CaC mol 10 31 . 1 O CaC g 128.10 O CaC mol mg 1000 g O CaC mg 2 . 167 - × = × ×
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 4-8 (a) 5 2 5 2 5 2 5 2 O P mmol 40 . 0 mol mmol 1000 O P g 94 . 141 O P mol mg 1000 g O P mg 57 = × × × (b) 2 2 2 2 CO mmol 6 . 293 mol mmol 1000 CO g 01 . 44 CO mol CO g 92 . 12 = × × (c) 3 3 3 3 NaHCO mmol 476 mol mmol 1000 NaHCO g 01 . 84 NaHCO mol NaHCO g 0 . 40 = × × (d) 4 4 4 4 4 4 4 4 PO MgNH mmol 2 . 6 mol mmol 1000 PO MgNH g 137.32 PO MgNH mol mg 1000 g PO MgNH mg 850 = × × × 4-9 (a) 4 4 3 4 3 KMnO mmol 50 . 6 L 00 . 2 mol mmol 1000 L KMnO mol 10 25 . 3 M KMnO 10 25 . 3 = × × × × - - (b) KSCN mmol 6 . 41 mL 750 mL 1000 L mol mmol 1000 L KSCN mol 0555 . 0 M KSCN 0555 . 0 = × × × (c) 4 3 4 4 4 4 CuSO mmol 10 47 . 8 mL 250 mL 1000 L mol mmol 1000 CuSO g 61 . 159 CuSO mol mg 1000 g L CuSO mg 41 . 5 CuSO ppm 41 . 5 - × = × × × × × (d) KCl mmol 6 . 1165 L 50 . 3 mol mmol 1000 L KCl mol 333 . 0 M KCl 333 . 0 = × ×
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 4-10 (a) 4 4 4 HClO mmol 0 . 56 mL 175 mL 1000 L mol mmol 1000 L HClO mol 320 . 0
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  • Fall '14
  • Mole, mol HCl, mol, mol Na, ml

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