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The mean indoor airborne chloroform (CHCl3) concentration in a room was determined to be 0.4 mg/m³. Use the following data: T = 293 K, P = 1 atmosphere, R = 82.05 × 10–6 atm. m³/mol-K, Atomic weights: C = 12, H = 1, Cl = 35.5. This concentration expressed in parts per billion (volume basis, ppbv) is equal to
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- 1.00 ppbv
- 0.20 ppbv
- 0.10 ppbv
- 0.08 ppbv
Correct Option: D
Molecular weight of CCl3 = 12 + 1 + 35.5 × 3 = 119.5 g
| No. of moles, n = | = | = 3.35 × 10-9 | ||
| 119.5 | 119.5 |
PV = nRT
| ∴ V = | = | = 0.08 × 10-9 = 0.08 ppbv | ||
| P | 1 |
