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5-day BOD and kg BOD in 3.2 ML

Short answer

Dilution P = 0.01. BOD₅ = (7.8 – 5.1)/0.01 = 270 mg/L. Mass in 3.2 million L = 270 mg/L × 3.2×10⁶ L = 864 kg.

2.5 ml of raw sewage is diluted to 250 ml. DO concentration of the diluted sample at the beginning was 7.8 mg/l and 51.0 [?] mg/l after 5 days of incubation at 20°C. Find 5-day BOD of raw sewage and kg BOD contained in 3.2 million liters of sewage.

The scanned value after incubation, "51.0", is not possible (it is more than the initial DO of 7.8 mg/L), so it is read as 5.1 mg/L.

Dilution

P=2.5250=0.01P=\frac{2.5}{250}=0.01

5-day BOD of raw sewage

BOD5=D1−D2P=7.8−5.10.01=270 mg/LBOD_5=\frac{D_1-D_2}{P}=\frac{7.8-5.1}{0.01}=270\ \text{mg/L}

Mass of BOD in 3.2 million litres

Mass=270 mg/L×3.2×106 L=864000000 mg=864 kg\text{Mass}=270\ \text{mg/L}\times3.2\times10^{6}\ \text{L}=864000000\ \text{mg}=864\ \text{kg}

Answer: BOD₅ = 270 mg/L; BOD in 3.2 ML = 864 kg.

✓ Answer checked by Claude Opus 5.5 (high effort)

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