CE3501 Environmental Engineering Fundamentals
Fall 2005
Problem Set 1 – Solutions
2-2. A typical loaf of bread contains 120 mg of sodium in each 1-ounce slice.
a) What is the concentration of sodium in the bread in ppm?
SOLUTION:
120mgNa 1oz 10−3 g 106 ppm
⋅
⋅
⋅
= 4233 ppm
1ozBread 28.35 g mg wt. fraction
b. Is the answer in units of ppmm or ppmv? Which makes sense and why?
ANSWER:
The answer is in units of ppmm. We do not know the volume of a slice of bread but we
do know its mass (1 oz) and we expressed the sodium concentration per unit mass of
bread (i.e., the mass of bread is in the denominator).
2-4. A water sample contains 10 mg NO3- /L. What is the concentration in
a) ppm?
SOLUTION:
10mgNO3− 10−3 g 1L
106 ppm
⋅
⋅ 3 ⋅
= 10 ppm
L
mg 10 g wt. fraction
b) moles/L
SOLUTION
10mgNO3− 10−3 g mole
moles
⋅
⋅
= 0.00016
L
mg 62 g
L
—
c) mg NO3 N/L
SOLUTION:
10mgNO3− 14mgN
mg ⋅ NO3− − N
⋅
=
2.26
L
62mgNO3−
L
d) ppb
SOLUTION:
Start with the answer to part a
103 ppb
= 104 ppb
10 ppm ⋅
ppm
2-21. An empty balloon is filled with exactly 10 g of nitrogren (N2) and 2 g of oxygen
(O2). The pressure in the room is 1.0 atm and the temperature is 25oC.
a) What is the oxygen concentration in the balloon in units of percent by volume?
SOLUTION:
Recall that volume fractions equal mole fractions. Therefore, since 10 g
equals (10/28) or 0.36 moles of nitrogen and 2 g of O2 equals 0.062 moles, the mole
fraction of oxygen is 0.062/(0.36+0.062) = 0.15 or 15%. (To convert to ppmv we would
simply multiply by 106 to arrive at 1.5x105 ppmv.)
b) What is the volume of the balloon after it is blown up?
SOLUTION:
Use the ideal gas law: PV = nRT
V=
nRT
L − atm ⎞
1
⎛
= ( 0.062 + 0.31moles ) ⋅ ⎜ 0.08205
= 9.0 L
⎟ ⋅ ( 25 + 273K ) ⋅
P
mole − K ⎠
1atm
⎝
2-23. “Clean” air might have a sulfur dioxide (SO2) concentration of 0.01 ppm while
“polluted” air might have a concentration of 2 ppm. Exposure to SO2 levels over
1 ppm may lead to breathing constrictions in the upper respiratory system.
Convert these two concentrations to μg/m3. Assume a temperature of 298K
(25oC).
SOLUTION:
(0.01 ppmv) * 10-6 (vol.frac./ppm) = 10-8 m3 SO2/m3 air = 10-8 moles SO2/mole air
−1
10−8
moleSO2
moleair
⋅ 64
g
moleSO2
⎛⎛
L − atm ⎞
⎞
0.08205
298K ) ⎟
(
⎜
⎟
⎜
L
μg ⎝
μg
mole − K ⎠
⎟ ⋅103 3 = 26.2 3
⋅106
⋅⎜
g ⎜
m
m
(1atm )
⎟
⎜
⎟
⎝
⎠
Repeating for the polluted value:
2 ppmv x 10-6 (vol.frac./ppm) = 2x10-6 moles SO2/mole air
−1
2 x10−6
moleSO2
moleair
⋅ 64
g
moleSO2
⎛⎛
L − atm ⎞
⎞
0.08205
298K ) ⎟
(
⎜
⎟
⎜
L
μg ⎝
μg
mole − K ⎠
⎟ ⋅103 3 = 5, 235 3
⋅106
⋅⎜
g ⎜
m
m
(1atm )
⎟
⎜
⎟
⎝
⎠