Fall, 2017

Homework Set 1.1 Text Solutions

Ch. 1: 1a, 5, 12, 16, 22, 26, 29, 34, 37 postponed to set 1.2

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





electric current


amount of substance
















5. The burning of fossil fuels by humans in 2012 introduced approximately 8 petagrams (Pg) of carbon per year into the atmosphere in the form of CO2.

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

kg of C yr-1 = (8 Pg C yr-1)(1015 g C/1 Pg C)(1 kg C/1000 g C) = 8 x 1012 kg C yr-1

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

kg CO2 yr-1 = (8 x 1012 kg C yr-1)(44.01 g CO2/mol CO2)(1 mol C/12.01 g C) = kg

CO2 yr-1 = 3 x 1013 kg CO2 yr-1

c) A metric ton is 1000 kg. How many metric tons of CO2 are placed in the atmosphere each year? There are 7 billion people on Earth. Find the per capita rate of CO2 production (tons of CO2 per person per year).

Metric tons person-1 yr-1 = (3 x 1013 kg CO2 yr-1)(1/7 x 109 persons)(1 metric ton/1000 kg)

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


12. 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


16. 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地)?

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


22. How many grams of perchloric acid, HClO4, are contained in 37.6 g of 70.5 wt% aqueous perchloric acid? How many grams of water are in the same solution?


g. HClO4 = (70.5 g HClO4/100 g sol地)(37.6 g sol地) = 26.5 g

g H2O = (100% - 70.5% g water)(37.6 g sol地) = 11.1 g



26. The concentration of sugar (glucose, C6H12O6) in human blood ranges from about 80 mg/dL before meals to 120 mg/ dL after eating. The abbreviation dL stands for deciliter = 0.1 L. Find the molarity of glucose before and after eating.

To complete the conversion we need the molecular weight of C6H12O6.

MW(C6H12O6) = 12.0*6 + 1.01*12 + 16.00*6 = 180.1 g/mol

Before eating: (80 mg glucose/100 mL)(1 g glucose/1000 mg glucose)(1 mol glucose/180.1 g glucose)(1000 mL/L) = 0.0044 M

After eating = 0.0067 M


29. It is recommended that drinking water contain 1.6 ppm fluoride (F-) for prevention of tooth decay. Consider a reservoir with a diameter of 4.50 x 102 m and a depth of 10.0 m. (The volume is pr2h, where r is the radius and h is the height.) How many grams of F- should be added to give 1.6 ppm? Fluoride is provided by hydrogen hexafluorosilicate, H2SiF6. How many grams of H2SiF6 contain this much F-?

We can approach this problem by using the reservoir dimensions to figure out the volume and mass of water. Then we can use the desired concentration to determine the mass of F- and H2SiF6 to be added.

Volume = (3.1416)(4.50 x 102 m/2)2(10.0 m) = 1.59 x 106 m3

Mass water = (1.59 x 106 m3)(100 cm/1 m)3(1.00 g water/cm3 water) = 1.59 x 1012 g water

1.6 ppm = 1.6 g F-/106 g water

Mass F- = (1.6 g F-/106 g water)(1.59 x 1012 g water) = 2.5 x 106 g F-

The mass of H2SiF6 is just the mass of F- divided by the mass fraction of F in H2SiF6

(= 6(19.00)/[2(1.01) + 28.08 + 6(19.00)] = 0.7911)

Mass of H2SiF6 = 2.5 x 106 g F-/0.7911 = 3.2 x 106 g


34. 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?

MconcVconc = MdilVdil and we are solving for Vconc

or Vconc = MdilVdil/Mconc = (1.00 M)(1.000 L)(1000 mL/L)/(18.0 M) = 55.6 mL

b) Calculate the density of 98.0 wt% H2SO4.

Density = g sol地/mL sol地

= (100 g sol地/98.0 g H2SO4)(98.1 g H2SO4/mol H2SO4)(18.0 mol H2SO4/L)(1 L/1000 mL) =

1.80 g/mL