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5-day BOD and kg BOD in 5 ML

Short answer

Dilution factor P = 2.5/250 = 0.01. BOD₅ = (8.0 – 5.0)/0.01 = 300 mg/L. Mass in 5 million L = 300 mg/L × 5×10⁶ L = 1500 kg.

2.5 ml of raw sewage is diluted to 250 ml. D.O. concentration of the diluted sample at the beginning was 8.0 mg/l and 5.0 [?] mg/l after 5 days of incubation at 20°C. Find 5-day B.O.D. of raw sewage and kg B.O.D. contained in 5 million liters of sewage.

The scanned value after incubation is read as 5.0 mg/L.

Dilution

P=volume of samplevolume of mixture=2.5250=0.01P=\frac{\text{volume of sample}}{\text{volume of mixture}}=\frac{2.5}{250}=0.01

5-day BOD of raw sewage

BOD5=D1−D2P=8.0−5.00.01=300 mg/LBOD_5=\frac{D_1-D_2}{P}=\frac{8.0-5.0}{0.01}=300\ \text{mg/L}

Mass of BOD in 5 million litres

Mass=300 mg/L×5×106 L=1500000000 mg=1500 kg\text{Mass}=300\ \text{mg/L}\times5\times10^{6}\ \text{L}=1500000000\ \text{mg}=1500\ \text{kg}

Answer: BOD₅ = 300 mg/L; BOD in 5 ML = 1500 kg.

✓ Answer checked by Claude Opus 5.5 (high effort)

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