Environmental Chemistry,
Department of Chemistry, University of Calgary
Practice problems 1 (Spring 2018)- Solutions
This material cannot be reproduced, redistributed or copied without the express consent of the
copyright holder.
1. Calculate the new volume of a quantity of gas occupying 15.6 L of volume at 0 oC and 1.10
atm pressure when conditions are changed to (a) 80 oC and 1.10 atm pressure and then to (b)
160oC and 0.90 atm pressure.
𝑃𝑉 = 𝑛𝑅𝑇
𝑃𝑉 (1𝑎𝑡𝑚)(15.6𝐿)
𝑛= = = 0.77 𝑚𝑜𝑙
𝑅𝑇 (0.08205Latm 𝐾 −1 𝑚𝑜𝑙 −1 )(273.15𝐾)
a)
𝑛𝑅𝑇 (0.77 𝑚𝑜𝑙)(0.08205Latm 𝐾 −1 𝑚𝑜𝑙 −1 )(273.15 + 80)𝐾
𝑉= = = 20.3𝐿
𝑃 1.10 𝑎𝑡𝑚
b)
𝑛𝑅𝑇 (0.77 𝑚𝑜𝑙)(0.08205Latm 𝐾 −1 𝑚𝑜𝑙 −1 )(2.73.15 + 160)𝐾
𝑉= = = 30.4𝐿
𝑃 0.9 𝑎𝑡𝑚
2. A sealed radioactive source used for physic demonstrations in 1940 contained 10 microcuries
(Microcurie -µCi- is a radioactivity measurement unit) of 60Co (cobalt- 60). Given a half life of
1900 days for 60Co, what would be the source strength in 1993?
Note: The reaction is a first order reaction, based on the unit for k value.
0.693 0.693 3.6×10−4
𝑡1/2 = 𝑘= = 𝑜𝑟 3.6 × 10−4 𝑑𝑎𝑦 −1
𝑘 1900 𝑑𝑎𝑦𝑠 𝑑𝑎𝑦
1993-1940=53 years
365 𝑑𝑎𝑦𝑠
53 years× 𝑦𝑒𝑎𝑟
= 19,345 days
3.6×10−4 𝑑𝑎𝑦 −1 )(19,345 𝑑𝑎𝑦𝑠)
𝐶𝑡 = 𝐶0 𝑒 −𝑘𝑡 = (10µCi)𝑒 −(10
Ct=8.6×10-3 µCi
, 3. An analytical chemist determines that a water sample contains 1.5 g/L of sulfate ion (SO42-). What is
the concentration in terms of
a. gram/L of sulfur (1:1 molar ratio of SO42- and S)
𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 32 𝑔 𝑠𝑢𝑙𝑓𝑢𝑟 𝑔𝑆
1.5 × = 0.5
𝐿 96 𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝐿
b. molar concentration of sulfate
𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 1 𝑚𝑜𝑙 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝑚𝑜𝑙
1.5 × = 0.016
𝐿 96 𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝐿
c. ppm (in dilute solutions=mg/L)
𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 1000 𝑚𝑔
1.5 × = 1500 𝑝𝑝𝑚
𝐿 𝑤𝑎𝑡𝑒𝑟 1𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒
4. Show that 10µg/m3 of sulfate aerosol is 10-11g/cm3
10µ𝑔 1𝑔 1𝑚3 1𝐿 𝑔
3
× 6
× × 3
= 10−11
𝑚 10 µ𝑔 1000𝐿 1000𝑐𝑚 𝑐𝑚3
5. In 1985, Huebert and coworkers (Tellus 40B, 260-269, 1988) measured average sulfate molar
mixing ratios in Beijing, China of 2050 pptv. Nowadays, sulfate concentrations are expressed in
μg m-3 of air. Convert Huebert et al.'s average concentrations to units of μg m-3. Assume a
pressure of 1.00 atm and a temperature of 15.0 °C.
P = 1.00 atm; T = (273.15+15) K; R = 0.08205 L atm mol-1 K-1; MW(Sulfate) = 96.0636 g/mol
PV = nRT
pptv: mole ratio: # moles of sulfate/# of moles of air=2050 mol sulfate/1012 mol air
Calculate the volume of 1012 moles of air using ideal gas law:
V=nRT/P=1012 mol×0.08205 L atm mol-1K-1× (273.15+ 15) K /1 atm =2.364×1013 L× 1m3/1000
L= 2.364×1010 m3
# of moles of sulfate: 2050 moles
g of sulfate: 2050mol×96.0636 g/mol=196930.38 g= 196930 μg
μg sulfate/m3 of air=196930.38×106 μg /2.364×1010 m3= 8.3μg/m3
or
Find the volume of 1 mol of air:
Department of Chemistry, University of Calgary
Practice problems 1 (Spring 2018)- Solutions
This material cannot be reproduced, redistributed or copied without the express consent of the
copyright holder.
1. Calculate the new volume of a quantity of gas occupying 15.6 L of volume at 0 oC and 1.10
atm pressure when conditions are changed to (a) 80 oC and 1.10 atm pressure and then to (b)
160oC and 0.90 atm pressure.
𝑃𝑉 = 𝑛𝑅𝑇
𝑃𝑉 (1𝑎𝑡𝑚)(15.6𝐿)
𝑛= = = 0.77 𝑚𝑜𝑙
𝑅𝑇 (0.08205Latm 𝐾 −1 𝑚𝑜𝑙 −1 )(273.15𝐾)
a)
𝑛𝑅𝑇 (0.77 𝑚𝑜𝑙)(0.08205Latm 𝐾 −1 𝑚𝑜𝑙 −1 )(273.15 + 80)𝐾
𝑉= = = 20.3𝐿
𝑃 1.10 𝑎𝑡𝑚
b)
𝑛𝑅𝑇 (0.77 𝑚𝑜𝑙)(0.08205Latm 𝐾 −1 𝑚𝑜𝑙 −1 )(2.73.15 + 160)𝐾
𝑉= = = 30.4𝐿
𝑃 0.9 𝑎𝑡𝑚
2. A sealed radioactive source used for physic demonstrations in 1940 contained 10 microcuries
(Microcurie -µCi- is a radioactivity measurement unit) of 60Co (cobalt- 60). Given a half life of
1900 days for 60Co, what would be the source strength in 1993?
Note: The reaction is a first order reaction, based on the unit for k value.
0.693 0.693 3.6×10−4
𝑡1/2 = 𝑘= = 𝑜𝑟 3.6 × 10−4 𝑑𝑎𝑦 −1
𝑘 1900 𝑑𝑎𝑦𝑠 𝑑𝑎𝑦
1993-1940=53 years
365 𝑑𝑎𝑦𝑠
53 years× 𝑦𝑒𝑎𝑟
= 19,345 days
3.6×10−4 𝑑𝑎𝑦 −1 )(19,345 𝑑𝑎𝑦𝑠)
𝐶𝑡 = 𝐶0 𝑒 −𝑘𝑡 = (10µCi)𝑒 −(10
Ct=8.6×10-3 µCi
, 3. An analytical chemist determines that a water sample contains 1.5 g/L of sulfate ion (SO42-). What is
the concentration in terms of
a. gram/L of sulfur (1:1 molar ratio of SO42- and S)
𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 32 𝑔 𝑠𝑢𝑙𝑓𝑢𝑟 𝑔𝑆
1.5 × = 0.5
𝐿 96 𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝐿
b. molar concentration of sulfate
𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 1 𝑚𝑜𝑙 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝑚𝑜𝑙
1.5 × = 0.016
𝐿 96 𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝐿
c. ppm (in dilute solutions=mg/L)
𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 1000 𝑚𝑔
1.5 × = 1500 𝑝𝑝𝑚
𝐿 𝑤𝑎𝑡𝑒𝑟 1𝑔 𝑠𝑢𝑙𝑓𝑎𝑡𝑒
4. Show that 10µg/m3 of sulfate aerosol is 10-11g/cm3
10µ𝑔 1𝑔 1𝑚3 1𝐿 𝑔
3
× 6
× × 3
= 10−11
𝑚 10 µ𝑔 1000𝐿 1000𝑐𝑚 𝑐𝑚3
5. In 1985, Huebert and coworkers (Tellus 40B, 260-269, 1988) measured average sulfate molar
mixing ratios in Beijing, China of 2050 pptv. Nowadays, sulfate concentrations are expressed in
μg m-3 of air. Convert Huebert et al.'s average concentrations to units of μg m-3. Assume a
pressure of 1.00 atm and a temperature of 15.0 °C.
P = 1.00 atm; T = (273.15+15) K; R = 0.08205 L atm mol-1 K-1; MW(Sulfate) = 96.0636 g/mol
PV = nRT
pptv: mole ratio: # moles of sulfate/# of moles of air=2050 mol sulfate/1012 mol air
Calculate the volume of 1012 moles of air using ideal gas law:
V=nRT/P=1012 mol×0.08205 L atm mol-1K-1× (273.15+ 15) K /1 atm =2.364×1013 L× 1m3/1000
L= 2.364×1010 m3
# of moles of sulfate: 2050 moles
g of sulfate: 2050mol×96.0636 g/mol=196930.38 g= 196930 μg
μg sulfate/m3 of air=196930.38×106 μg /2.364×1010 m3= 8.3μg/m3
or
Find the volume of 1 mol of air:
