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 CO_{2}.
a) How many kg of C were placed in the atmosphere each year?
kg of C
yr^{-1} = (8 Pg C yr^{-1})(10^{15}
g C/1 Pg C)(1 kg C/1000 g C) = 8 x 10^{12} kg C yr^{-1}
b) How many kg of CO_{2} were placed in the atmosphere each year?
kg CO_{2} yr^{-1} = (8 x 10^{12} kg C yr^{-1})(44.01
g CO_{2}/mol CO_{2})(1 mol C/12.01 g C) = kg
CO_{2} yr^{-1} = 3 x 10^{13} kg CO_{2} yr^{-1}
c) A metric ton is 1000 kg. How many metric tons of CO_{2} are placed in the atmosphere each year? There are 7 billion people on Earth. Find the per capita rate of CO_{2} production (tons of CO_{2} per person per year).
Metric tons person^{-1} yr^{-1} = (3 x 10^{13}
kg CO_{2} yr^{-1})(1/7 x 10^{9}
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 km^{2})
metric tons Pb yr^{-1} = 6 metric tons Pb yr^{-1}
16. How many grams of methanol (CH_{3}OH, FM 32.04) are contained in 0.100 L of 1.71 M aqueous methanol (i.e. 1.71 mol CH_{3}OH/L sol’n)?
Mass CH_{3}OH = (1.71 mol CH_{3}OH/L sol’n)(32.04 g CH_{3}OH /mol CH_{3}OH)(0.100 L) = 5.48 g
22. How many grams of perchloric acid, HClO_{4}, are contained in 37.6 g of 70.5 wt% aqueous perchloric acid? How many grams of water are in the same solution?
g. HClO_{4} = (70.5 g HClO_{4}/100 g sol’n)(37.6 g sol’n)
= 26.5 g
g H_{2}O = (100% - 70.5% g water)(37.6 g sol’n)
= 11.1 g
26. The concentration of sugar (glucose, C_{6}H_{12}O_{6}) 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 C_{6}H_{12}O_{6}.
MW(C_{6}H_{12}O_{6}) = 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 10^{2} m and a depth of 10.0 m. (The volume is pr^{2}h,
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,
H_{2}SiF_{6}. How many
grams of H_{2}SiF_{6} 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 H_{2}SiF_{6} to be added.
Volume = (3.1416)(4.50 x 10^{2} m/2)^{2}(10.0
m) = 1.59 x 10^{6} m^{3}
Mass water = (1.59 x 10^{6} m^{3})(100
cm/1 m)^{3}(1.00 g water/cm^{3} water) = 1.59 x 10^{12}
g water
1.6 ppm = 1.6 g F^{-}/10^{6} g water
Mass F^{-} = (1.6 g F^{-}/10^{6} g water)(1.59 x 10^{12} g water) = 2.5 x 10^{6} g F^{-}
The mass of H_{2}SiF_{6} is just the mass of F^{-}
divided by the mass fraction of F in H_{2}SiF_{6}
(= 6(19.00)/[2(1.01) + 28.08 + 6(19.00)] =
0.7911)
Mass of H_{2}SiF_{6} = 2.5 x 10^{6} g F^{-}/0.7911 = 3.2 x 10^{6} g
34. A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% H_{2}SO_{4}, 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_{conc}V_{conc} = M_{dil}V_{dil}
and we are solving for V_{conc}
or V_{conc} = M_{dil}V_{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% H_{2}SO_{4}.
Density = g sol’n/mL sol’n
= (100 g sol’n/98.0 g H_{2}SO_{4})(98.1 g H_{2}SO_{4}/mol
H_{2}SO_{4})(18.0 mol H_{2}SO_{4}/L)(1
L/1000 mL) =
1.80 g/mL