Download Tutorial # 2 2006

SCHOOL OF CHEMISTRY, UKZN.APCH211: ENV. CHEMISTRY- Tutorial Sheet-2
Lecturer: Prof. S.B. Jonnalagadda
1. Convert a concentration of 40 ppb of ozone into (a) the number of molecules per cm3, (b) the number of
moles per liter and (c). mg per m3. Assume the air mass temperature is 27oC and it’s total pressure is 0.95
atmospheres. (Ans: a = 9.29 x 1011 molecules per cm3, b = 1.544 x 10-9 M, c = 74.1 x10-3 mg m-3)
2. Convert 320 micrograms/m3 to ppb scale, if the pollutant is SO2, the total pressure is 1.0 atm and the
temperature is 27oC. (Ans: 123 ppb)
3. Convert a concentration of 32 ppb for any pollutant to its value on a. the ppm scale, b. the molecules per
cm3 scale and c. moles per liter scale. Assume 25oC and a total pressure of 1.0 atm.
(Ans: a = 0.032 ppm, b = 7.9 x 1011 molecules per cm3, c = 1.31 x 10-9 moles per litre)
4. The average outdoor concentration of carbon monoxide, CO, is about 1000 µg/m3. What is the
concentration expressed a. on the ppm scale and b. on the molecules per cm3 scale? Assume that the
outdoor temperature is 17oC and that the total air pressure is 1.04 atmospheres. (Ans: a: 0.817 ppm; b =
2.15 x 1013 moles cm-3)
5. The day time concentration of OH* in continental air averages 7.5 x 106 molecules per cm3. Calculate it’s
molar concentration and it’s concentration in ppt, assuming that the total air pressure is 1.0 atm and the
temperature is 15oC. (Ans: a =1.25 x 10-14 M and b = 0.3 ppt)
6. Deduce the balanced reaction in which ammonia reacts with nitrogen dioxide to produce molecular
nitrogen and water. Calculate the mass of ammonia that is required to react with 1000 L of air containing
10 ppm of NO2. Assume the air at 27oC and 1.0 atmosphere. (Ans: = 9.24 x 10-3 g)
7. In the flue-gas desulfurization process, the removal of SO2 using Calcium carbonate is represented by the
following equation:
CaCO3 + SO2 → CaSO3 + CO2 and 2CaSO3 + O2 → 2 CaSO4
What mass of calcium carbonate is required to react with the sulfur dioxide that is produced by burning one
tonn (1000 kg) of coal that contains 5.0 % sulfur by mass? (Ans: 155.9 kg)
8. Decide whether the TLV is exceeded in each of the following cases.(Assume 1 atm pressure and
temperature 288 K)
(a) SO2 (1.6 ppm, TLV = 5 mg m-3) and chlorine (0.9 ppm, TLV = 1.5 mg m-3)
(b) benzene (1.5 ppm, TLV = 30 mg m-3) and toluene (85 ppm, TLV = 375 mg m-3)
9. In a polluted premises of volume 200 m3, 15 mg of ozone (TLV = 0.2 mg m-3), 140 mg of chlorine (TLV
= 1.5 mg m-3), and 4000 mg of carbon monoxide (TLV = 55 mg m-3), were detected in the air. Assuming,
the pollutants have additive effect, estimate whether pollutants exceed the combined threshold value.
(Ans: 1.21, exceeds TLV)
10.(a) Decide whether the TLV is exceeded in each of the following cases.(Assume 1 atm pressure and
temperature 288 K) Heptane (180 ppm, TLV = 1600 mg m-3), octane (100 ppm, TLV = 1450 mg m-3) and
nonane (25 ppm, TLV = 1650 mg m-3).
Ans: Hp = 763.6 mg m-3; Oc = 483.5 mg m-3; No =135.8 mg m-3 (All with in TLV)
(b). Stating any assumptions you make, derive the composite relationship given, the compositions of the
liquids by weight are heptane (20%), octane (30%) and nonane (50%). (Ans: 1575 µg m-3)
11. Deduce the formula for the compounds with the following code numbers:
a. 12
b. 113 c, 123 d. 134 e. 22
(Ans: a = CF2Cl2, b = C2F3Cl3, c = C2HF3Cl2, d = C2H2F4, e = CHF2Cl
12. Deduce the code numbers for each of the following compounds:
a. CH3CCl3 b. CCl4 c. CH3CFCl2 d. CHCl2 CHCl2. (Ans: a. = 140, b =10 c =141, d = 130)
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SCHOOL OF CHEMISTRY, UKZN.
APCH211: ENVIRONMENTAL CHEMISTRY
Tutorial Sheet-2
1. Convert a concentration of 40 ppb of ozone into (a) the number of
molecules per cm3, (b) the number of moles per liter and (c) micrograms per
m3. Assume the air mass temperature is 27oC and it’s total pressure is 0.95
atmospheres. (Ans: a = 9.29 x 1011 molecules per cm3, b = 1.544 x 10-9 M, c
= 74.1 x 10-3µg m-3)
a. n = PV/RT
1 L = 1 dm3 = (0.1 m)3
= 0.95 atm x 1 m3/ {(0.082 L atm mole-1 K-1) x 10-3 m3 x 300 K}
= 38.6 moles
No moles of ozone = 40 x 10-9 x 38.6 = 1.544 x 10-6 moles per m3
1 cm3 = (0.01m)3 = 10-6 m3
No moles of ozone = 1.544 x 10-6 moles m-3 /106 cm3 m-3
= 1.544 x 10-12 moles cm-3
No molecules of ozone = 1.544 x 10-12 moles cm-3 x 6.02 x1023
= 9.29 x 1011 molecules per cm3
b. No moles of ozone/L = 1.544 x 10-12 moles cm-3 x 1000 cm3 per L
= 1.544 x 10-9 M
c. Mass of ozone = 1.544 x 10-6 moles per m3 x 48 g
= 74.1 x 10-6 g = 74.1 x 10-3 mg m-3.
2. Convert 320 micrograms/m3 to ppb scale, if the pollutant is SO2, the total
pressure is 1.0 atm and the temperature is 27oC. (Ans: 123 ppb)
No moles of SO2 = mass / molar mass
= 320 x 10-6 g /64.1 g = 4.99 x 10-6 moles
No molecules of SO2 = 4.99 x 10-6 moles x 6.02 x 1023 molecules/mole =
3.01 x 1018 Using the ideal gas equation, the volume of air can be
converted into number of moles and molecules.
n = PV/RT 1 L = 1 dm3 = (0.1 m)3
= 1 atm x 1 m3/ {(0.082 L atm mole-1 K-1) x 10-3 m3 x 300 K}
= 40.7 moles
No. molecules = 40.7 x 6.02 x 1023 = 2.45 x 1025 molecules
Thus concentration of SO2 = 3.01 x 1018/2.45 x 1025
= 123 x 10-9 = 123 ppb
3. Convert a concentration of 32 ppb for any pollutant to its value on
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a. the ppm scale, b. the molecules per cm3 scale and c. moles per liter scale.
Assume 25oC and a total pressure of 1.0 atm.(Ans: a = 0.032 ppm, b = 7.9 x
1011 molecules per cm3, c = 1.31 x 10-9 moles per litre)
a. 32 ppb = 0.032 ppm
b. Using the ideal gas equation, the volume of air can be converted into
number of moles and molecules.
n = PV/RT 1 L = 1 dm3 = (0.1 m)3
= 1 atm x 1 m3/ {(0.082 L atm mole-1 K-1) x 10-3 m3 x 298 K}
= 40.9 moles
No moles of pollutant = 32 x 10-9 x 40.9 = 1.31 x 10-6 moles per m3 =
1 cm3 = (0.01m)3 = 10-6 m3
No moles of pollutant = 1.31 x 10-6 moles m-3 /106 cm3 m-3
= 1.31 x 10-12 moles cm-3
No molecules =1.31 x 10-12 moles cm-3 x 6.02 x 1023 molecules/ mole
= 7.9 x 1011 molecules per cm3
c. No moles per litre = 1.31 x 10-12 x 1000 cm3 L-1
= 1.31 x 10-9 moles per litre.
4. The average outdoor concentration of carbon monoxide, CO, is about
1000 µg/m3. What is the concentration expressed a. on the ppm scale and b.
on the molecules per cm3 scale? Assume that the outdoor temperature is
17oC and that the total air pressure is 1.04 atmospheres. (Ans: a: 0.817 ppm;
b = 2.15 x 1013 moles cm-3)
No moles = 1000 x10-6 g/ 28 g = 35.7 x 10-6 moles
= 35.7 x 10-6 moles x 6.02 x 1023 molecules/mole
= 2.15 x 1019 molecules per m3
1 cm3 = (0.01m)3 = 10-6 m3
No moles of CO = 2.15 x 1019 moles m-3 /106 cm3 m-3
= 2.15 x 1013 moles cm-3
n = PV/RT 1 L = 1 dm3 = (0.1 m)3
= 1.04 atm x 1 m3/ {(0.082 L atm mole-1 K-1) x 10-3 m3 x 290 K}
= 43.7 moles
Concentration = Moles of CO/Moles of air = 35.7 x 10-6/43.7
= 817 ppb = 0.817 ppm
5. The day time concentration of OH* in continental air averages 7.5 x 106
molecules per cm3. Calculate it’s molar concentration and it’s concentration
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in ppt, assuming that the total air pressure is 1.0 atm and the temperature is
15oC. (Ans: a =1.25 x 10-14 M and b = 0.3 ppt)
1 cm3 = (0.1 dm)3 = 10-3 dm3
No molecules of OH* = 7.5 x 106 molecules cm-3 x 103 cm3 dm-3
= 7.5 x 109 molecules dm-3
No moles of OH* = 7.5 x 109 molecules dm-3 / 6.02 x 1023 molecules/mole
= 1.25 x 10-14 moles per liter = 1.25 x 10-14 M
n = PV/RT 1 L = 1 dm3 = (10cm)3 = 103 cm3 dm-3
= 1.0 atm x 1 dm3 / {(0.082 L atm mole-1 K-1) x 288 K}
= 0.0423 moles
Concentration = Moles of OH*/Moles of air = 1.25 x 10-14/0.0423
= 29.55 x 10-14= 0.296 10-12 = 0.3 ppt
6. Deduce the balanced reaction in which ammonia reacts with nitrogen
dioxide to produce molecular nitrogen and water. Calculate the mass of
ammonia that is required to react with 1000 L of air containing 10 ppm of
NO2. Assume the air at 27oC and 1.0 atmosphere. (Ans: = 9.24 x 10-3 g)
a. 8NH3 + 6NO2 = 7N2 + 12 H2O
b. Using the ideal gas equation, the volume of air can be converted into
number of moles and molecules.
n = PV/RT V =1000 L = 1 m3
= 1 atm x 1 m3/ {(0.082 L atm mole-1 K-1) x 10-3 m3 x 300 K}
= 40.7 moles
No moles of NO2 = 10 x 10-6 x 40.7 = 407 x 10-6 moles per m3
1 mole of NO2 need 8/6 moles of NH3
Mass of ammonia needed = 4.07 x 10-4 x (8/6) x 17.02 g/mole
= 9.24 x 10-3 g
7. In the flue-gas desulfurization process, the removal of SO2 using Calcium
carbonate is represented by the following equation.
CaCO3 + SO2 → CaSO3 + CO2 and 2CaSO3 + O2 → 2 CaSO4
What mass of calcium carbonate is required to react with the sulfur dioxide
that is produced by burning one tonne (1000 kg) of coal that contains 5.0 %
sulfur by mass? (Ans: 155.9 kg)
No moles of SO2 produced = Mass of sulfur/Atomic mass of S
= 1000 kg x 103 g/kg x (5/100) /32.1 g = 1558 moles
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Mass of Calcium carbonate required = 1558 moles x 100.08 g/mole
= 155.9 kg
8. Decide whether the TLV is exceeded in each of the following
cases.(Assume 1 atm pressure and temperature 288 K)
(a) SO2 (1.6 ppm, TLV = 5 mg m-3) and Cl2 (0.9 ppm, TLV =1.5 mg m-3)
(b) benzene (1.5 ppm, TLV = 30 mg m-3) and toluene (85 ppm, TLV = 375
mg m-3)
(c) heptane (180 ppm, TLV = 1600 mg m-3), octane (100 ppm, TLV = 1450
mg m-3) and nonane (25 ppm, TLV = 1650 mg m-3).
(a) SO2 (1.6 ppm) and chlorine (0.9 ppm)
a. SO2 (TLV = 5 mg m-3)
n = PV/RT
1 L = 1 dm3 = (0.1 m)3
= 1 atm x 1 m3/ {(0.082 L atm mole-1 K-1) x 10-3 m3 x 288 K}
= 42.34 moles of air
No moles of SO2 = 1.6 x 10-6 x 42.34 = 6.77 x 10-5 moles m-3
Mass of SO2 = 6.77 x 10-5 moles per m3 x 64 g mole-1
= 4.333 x 10-3 g = 4.333 mg m-3.(With in TLV)
Cl2 (TLV = 1.5 mg m-3)
No moles = 0.9 x 10-6 x 42.34 = 3.81 x 10-5 moles m-3
Mass of Cl2 = 6.77 x 10-5 moles per m3 x 70.9 g mole-1
= 2.702 x 10-3 g = 2.702 mg m-3.(Exceeds TLV)
(b) benzene (1.5 ppm) and toluene (85 ppm)
Benzene, C6H6 (TLV = 30 mg m-3)
n = 42.34 moles of air
No moles of benzene = 1.5 x 10-6 x 42.34 = 6.35 x 10-5 moles m-3
Mass of benzene = 6.35 x 10-5 moles per m3 x 78.1 g mole-1
= 4.960 x 10-3 g = 4.96 mg m-3. (With in TLV)
Toluene, C7H8 (TLV = 375 mg m-3)
No moles = 85 x 10-6 x 42.34 = 3599 x 10-6 moles m-3
Mass of toluene = 3599 x 10-6 moles per m3 x 92.1 g mole-1
= 331468 x 10-6 g = 331.5 mg m-3 (With in TLV)
9. In a polluted premises of volume 200 m3, 15 mg of ozone (TLV = 0.2 mg
m-3), 140 mg of chlorine (TLV = 1.5 mg m-3),and 4000 mg of CO (TLV = 55
mg m-3), were detected in the air. Assuming, the pollutants have additive
effect, estimate whether pollutants exceed the combined threshold value.
(Ans: 1.21, exceeds TLV)
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Concn. of O3 = 15/200 = 0.75 mg m-3
Fraction of Ozone = Concn/TLV = 0.075/0.2 = 0.375
Concn. of Cl2 = 140/200 = 0.75 mg m-3
Fraction of Cl2 = Concn/TLV = 0.07/1.5 = 0.467
Concn. of CO = 4000/200 = 20 mg m-3
Fraction of CO = Concn/TLV = 20/55 = 0.364
c1
TLV1
+
c2
TLV2
+
c3
+
TLV3
…>1
Combined value = 0.375 + 0.467 + 0.364 = 1.21 which is greater than 1.
The concentrations exceed the combined TLV.
10.a Decide whether the TLV is exceeded in each of the following
cases.(Assume 1 atm pressure and temperature 288 K)
Heptane (180 ppm, TLV = 1600 mg m-3), octane (100 ppm, TLV = 1450 mg
m-3) and nonane (25 ppm, TLV = 1650 mg m-3).
Ans: Hp = 763.6 mg m-3; Oc = 483.5 mg m-3; No =135.8 mg m-3 (All with in
TLV)
b. Stating any assumptions you make, derive the composite relationship
given,
TLVmix =
{
f1
TLV1
+
f2
TLV2
+
f3
+
TLV3
… }-1
The compositions of the liquids are given by weight, heptane (20%), octane
(30 %) and nonane (50 %). (Ans: 1575 µg m-3)
a. Heptane (180 ppm, TLV = 1600 mg m-3)
n = 42.34 moles of air
No moles of heptane = 180 x 10-6 x 42.34 = 7621 x 10-6 moles m-3
Mass of heptane = 7621 x 10-5 moles per m3 x 100.2 g mole-1
= 763644 x 10-6 g = 763.6 mg m-3 (With in TLV)
Octane, C8H18 (100 ppm, TLV = 1450 mg m-3)
No moles = 100 x 10-6 x 42.34 = 4234 x 10-6 moles m-3
Mass of octane = 4234 x 10-6 moles per m3 x 114.2 g mole-1
= 483.5 x 10-6 g = 483.5 mg m-3 (With in TLV)
Nonane, C9H20 (25 ppm, TLV = 1650 mg m-3)
No moles = 25 x 10-6 x 42.34 = 1059 x 10-6 moles m-3
Mass of nonane = 1059 x 10-6 moles per m3 x 128.3 g mole-1
= 135806 x 10-6 g = 135.8 mg m-3 (With in TLV).
b.
TLVmix =
{ 0.2
1600
+
0.3 + 0.5 }-1
1450
1650
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= {1.25 x10-4 + 2.07 x10-4 + 3.03 x10-4 }-1
= (6.35 x 10-4)-1 = 1575 mg m-3
TLV for composite mixture equals to 1575 mg m-3
11. Deduce the formula for the compounds with the following code
numbers: a. 12
b. 113 c, 123
d. 134 e. 22
(Ans: a = CF2Cl2, b = C2F3Cl3, c = C2HF3Cl2, d = C2H2F4, e = CHF2Cl
a. CFC-12
12 + 90 = 102; C = 1, H = 0 and F = 2
Chlorine atoms = 2 + 2 - 2 = 2
∴CFC-12 = CF2Cl2
b. CFC-113
113 + 90 = 203; C = 2, H = 0 and F = 3
Chlorine atoms = 4 + 2 - 3 = 3
∴CFC-113 = C2F3Cl3 = CF2ClCFCl2
c. 123
123 + 90 = 213; C = 2, H = 1 and F = 3
Chlorine atoms = 4 + 2 - 4 = 2
∴CFC-113 = C2HF3Cl2 = CHFCl CF2Cl
d. 134
134 + 90 = 224; C = 2, H = 2 and F = 4
Chlorine atoms = 4 + 2 - 6 = 0
∴CFC-113 = C2H2F4 = CHF2 CHF2
e. 22
12 + 90 = 112; C = 1, H = 1 and F = 2
Chlorine atoms = 2 + 2 - 3 = 1
∴CFC- 22= CHF2Cl (Hydrochlorofluorocarbon)
12. Deduce the code numbers for each of the following compounds:
a. CH3CCl3 b. CCl4 c. CH3CFCl2
d. CHCl2 CHCl2
(Ans: a. = 140, b =10 c =141, d = 130)
a. 230-90 =CFC-140 b. 100-90 = CFC-10 c. 231-90 =CFC-141 d. 220-90
=CFC-130
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