Page 30 - Materials Chemistry, Second Edition

Site Assessment and Remedial Investigation 13

to more benzene? Note that the typical water intake rate is 2.0 L/day and the
air inhalation rate is 15.2 m /day for an adult [13].
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Solution:
(a) Benzene ingested daily = (2 L/day)(10 × 10 mg/L) = 0.02 mg/day
−3
(b) Benzene inhaled daily:
MW of benzene (C H ) = (12)(6) + (1)(6) = 78 g/mole
6
6
At T = 20°C and P = 1 atm,
10 ppbV benzene = (10 × 10 )(78/24.05) mg/m = 0.0324 mg/m 3

3
−3
Benzene inhaled daily = (15.2 m /day)(0.0324 mg/m ) = 0.49 mg/

3
3
day
∴ The inhalation system is exposed to more benzene.
Discussion:
1. 1 ppb = 0.001 ppm
2. The pressure was not specified, and P = 1 atm was used
(a reasonable assumption).

Example 2.4: Mass and Concentration Relationship
A glass bottle containing 900 mL of methylene chloride (CH Cl , specific
2
2
gravity = 1.335) was accidentally left uncapped over a weekend in a poorly
ventilated room (5 m × 6 m × 3.6 m). On the following Monday, it was found
that two-thirds of methylene chloride had volatilized from the bottle.
For the worst-case scenario (i.e., all the volatilized methylene chloride
stayed in the room, with no air exchange with the outside), would the concen-
tration in the air exceed the Occupational Safety and Health Administration
(OSHA’s) eight-hour time-weighted average (TWA) permissible exposure
limit (PEL) of 25 ppmV and the short-term exposure limit (STEL) of 125
ppmV?

Solution:
(a) Mass of methylene chloride volatilized = (liquid volume) ×
(density)
= [(2/3)(900 mL)](1.335 g/mL)
= (600)(1.335) = 801 g = 8.01 × 10 mg
5
(b) Vapor concentration in mass/vol = (mass) ÷ (volume)
= (8.01 × 10 mg) ÷ [(5 m)(6 m)(3.6 m)]
5
= (8.01 × 10 ) ÷ (108) = 7,417 mg/m 3
5