CHEMISTRY 31

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

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. 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’n)?

Mass CH₃OH = (1.71 mol CH₃OH/L sol’n)(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’n)(37.6 g sol’n) = 26.5 g

g H₂O = (100% - 70.5% g water)(37.6 g sol’n) = 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?

M_concV_conc = M_dilV_dil and we are solving for V_conc

or V_conc = M_dilV_dil/M_conc = (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’n/mL sol’n

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

1.80 g/mL