CHEMISTRY 31

Homework Set 1.1 Solutions

Chapter Problems Date to finish

Set 1.1 Sept. 5

Ch. 1 1a, 5, 11, 13, 15, 20, 24,

32, 35, 37, 42, 46

1. a) List the SI units of length, mass, time, electric current, temperature and amount of substance

Measure	length	mass	time	electric current	temperature	amount of substance
Unit	meter	kilogram	second	ampere	kelvin	mole
Abbreviation	m	kg	s	A	K	mol

5. During the 1980s, the average emission of carbon from burning fossil fuels on Earth was 5.4 petagrams (Pg) of carbon per year in the form of CO₂.

a) How many kg of C were placed in the atmosphere each year?

kg of C yr^-1 = (5.4 Pg C yr^-1)(10¹⁵ g C/1 Pg C)(1 kg C/1000 g C) = 5.4 x 10¹² kg C yr^-1

b) How many kg of CO₂ were placed in the atmosphere each year?

kg CO₂ yr^-1 = (5.4 x 10¹² kg C yr^-1)(44.01 g CO₂/mol CO₂)(1 mol C/12.01 g C) = kg

CO₂ yr^-1 = 2.0 x 10¹³ kg CO₂ yr^-1

c) A metric ton is 1000 kg. How many metric tons of CO₂ were placed in the atmosphere each year? If there were 5 billion people on Earth, how many tons of CO₂ were produced for each person?

Metric tons person^-1 yr^-1 = (2.0 x 10¹³ kg CO₂ yr^-1)(1/5 x 10⁹ persons)(1 metric ton/1000 kg)

Metric tons person^-1 yr^-1 = 4 metric tons person^-1 yr^-1

11. Dust falls on Chicago at a rate of 65 mg m^-2 day. Major metallic elements in the dust include Al, Mg, Cu, Zn, Mn, and Pb. Pb accumulates at a rate of 0.03 mg m^-2 day^-1. How many metric tons (1 metric ton = 1000 kg) of Pb fall on the 535 square kilometers of Chicago in a year?

metric tons Pb yr^-1 = (0.03 mg Pb m^-2 day^-1)(1 g/1000 mg)(1 kg/1000 g)(1 metric ton/1000 kg)

(365 day/1 yr)(1000 m/1 km)² (535 km²)

metric tons Pb yr^-1 = 6 metric tons Pb yr^-1

13. Why is it more accurate to say that the concentration of a solution of acetic acid is 0.01 F rather than 0.01 M? (Despite this distinction, we will usually write 0.01M).

Because acetic acid in solution will exist as a mixture of acetic acid and dissociated acetic acid (or acetate), it is more accurate to use F (formal) where the combined concentration of acetic acid and acetate is equal to 0.01 F.

15. How many grams of methanol (CH₃OH, FM 32.04) are contained in 0.100 L of 1.71 M aqueous methanol (i.e. 1.71 mol CH₃OH/L sol’n)?

Mass CH₃OH = (1.71 mol CH₃OH/L sol’n)(32.04 g CH₃OH /mol CH₃OH)(0.100 L) = 5.48 g

20. The density of 70.5 wt % perchloric acid is 1.67 g/mL. Recall that grams refers to grams of solution.

a) How many grams of solution are in 1.00 L?

mass solution = (1.67 g/mL)*(1.00 L)*(1000 mL/L) = 1670 g.

b) How many grams of HClO₄ are in 1.00 L?

mass HClO₄ = (1670 g)*(0.705 g HClO₄/g sol'n) = 1.18 x 10³ g.

c) How many moles of HClO₄ are in 1.00 L?

moles HClO₄ = (1.18 x 10³)/((1 + 35.5 + 16*4) g/mol) = 11.7 moles

24. An aqueous solution of antifreeze contains 6.067 M ethylene glycol (HOCH₂CH₂OH, FM 62.07) and has a density of 1.046 g/mL.

a) Find the mass of 1.000 L of this solution and the number of grams of ethlyene glycol in this solution.

Mass solution = (1.000 L)(1.046 g/mL)(1000 mL/L) = 1046 g.

Mass ethylene glycol = (6.067 mol/L)(1.000 L)(62.07) = 376.6 g

b) Find the molality of ethlyene glycol in this solution.

Molality = mol/kg solvent. 1.000 L solution means 6.067 moles

and 1046 g solution – 376.6 g solute = 669.4 g = 0.669 kg

Molality = 6.067 mol/0.6694 kg = 9.06 m

32. A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% H₂SO₄, has a concentration of 18.0 M.

a) How many milliliters of reagent should be diluted to 1.000 L to give 1.00 M H₂SO₄?

(18.0 mol L^-1)V = (1.00 mol L^-1)(1.000 L) V = (0.05556 L)(1000 mL/L)

V = 55.6 mL

b) Calculate the density of 98.0 wt% H₂SO₄.

density = mass/volume. 1.000 L has 18.0 moles

mass in 1.000 L = (18.0 mol)(98.08 g mol^-1) = 1765 g

Density = 1765 g/1000 mL = 1.77 g mL^-1. (slightly different answer in book)

The quiz will not be on problems 35-46

35. How many grams of 0.491 wt % aqueous HF are required to provide a 50% excess to react with 25.0 mL of 0.0236 M Th⁴⁺ by the reaction Th⁴⁺ + 4F^- => ThF₄ (s)?

mmoles Th⁴⁺ = (23.6 mmol/L)*(25.0 mL)*(10^-3 L/mL) = 0.590 mmol

mmol F^- required = 0.59*(4 mol F^-/1 mol Th⁴⁺) = 2.36 mmol

mmol F^- required for 50% excess = 1.5*(2.36 mmol) = 3.54 mmol

g HF sol'n needed = (20.0 mg HF/mmol HF)*(3.54 mmol HF)*(10^-3g/mg)/(0.00491 mgHF/mg sol'n) = 14.4 g.

37. Distinguish between the terms end point and equivalence point.

Equivalence point refers to the point in the titration when the reagents have been added at the exact stoichiometric ratio. End point refers to the point when the equivalence point is detected.

42. How many milliliters of 0.100 M KI are needed to react with 40.0 mL of 0.0400 M Hg₂(NO₃)₂ if the reaction is Hg₂²⁺ + 2I^- → Hg₂I₂(s)?

note: n = number of moles

n(KI)/n(Hg₂(NO₃)₂) = 2/1 n(Hg₂(NO₃)₂) = (40.0 mL)(0.0400 mmol/mL) = 1.60 mmol

n(KI) = 2n(Hg₂(NO₃)₂) = 3.20 mmol = [KI]V = 0.100 mmol/mLV

V = 32.0 mL

46. Limestone consists mainly of the mineral calcite, CaCO₃ (s) (FW = 100.087 g/mol). The carbonate content of 0.5413 g of powdered limestone was measured by suspending the powder in water, adding 10.00 mL of 1.396 M HCl, and heating to dissolve the solid and expel CO₂:

CaCO₃ (s) + 2H⁺ => Ca²⁺ + CO₂ (g) + H₂O

The excess acid requires 39.96 mL of 0.1004 M NaOH for complete titration to a phenolphthalein endpoint. Find the weight percent of calcite in the limestone.

n(HCl)_added = n(HCl)_excess + n(HCl)_reacted and n(HCl)_reacted = n(NaOH)

n(HCl)_reacted/n(CaCO₃) = 2/1 or

n(CaCO₃) = 0.5*n(HCl)_reacted = 0.5*( n(HCl)_added - n(HCl)_reacted)

n(CaCO₃) = 0.5*[(10.00 mL*1.396 mmol/mL) - (39.96mL*0.1004 mmol/mL)] = 4.974 mmol

mass percent calcite = 4.974 mmol*100.087mg/mmol*0.001 g/mg*100.09/0.5413

mass percent calcite = 92.0%