---
title: "5-day BOD and kg BOD in 5 ML"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-dilution-2p5-in-250-8mg"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-dilution-2p5-in-250-8mg
site: BCT Engineering
---

# 5-day BOD and kg BOD in 5 ML

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2071 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-dilution-2p5-in-250-8mg
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

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.

## 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.

## Answer

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

#### Dilution
$$
P=\frac{\text{volume of sample}}{\text{volume of mixture}}=\frac{2.5}{250}=0.01
$$

#### 5-day BOD of raw sewage
$$
BOD_5=\frac{D_1-D_2}{P}=\frac{8.0-5.0}{0.01}=300\ \text{mg/L}
$$

#### Mass of BOD in 5 million litres
$$
\text{Mass}=300\ \text{mg/L}\times5\times10^{6}\ \text{L}=1500000000\ \text{mg}=1500\ \text{kg}
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is P for 2.5 mL in 250 mL?
   A. 0.01
   B. 0.1
   C. 100
   D. 0.025
   Answer: A. P = 2.5/250 = 0.01. (Key point: Dilution fraction)

2. What is BOD5 of raw sewage?
   A. 3 mg/L
   B. (8.0 - 5.0)/0.01 = 300 mg/L
   C. (8.0 + 5.0)/0.01 = 1300 mg/L
   D. 3 x 0.01 = 0.03 mg/L
   Answer: B. BOD5 = (D1 - D2)/P. (Key point: BOD formula)

3. How is the mass in 5 million litres found?
   A. 300 x 5 = 1500 mg
   B. 300/5 million
   C. Concentration x volume: 300 mg/L x 5 x 10^6 L = 1500 kg
   D. 300 x 5 x 10^6 = 1500 g
   Answer: C. 300 mg/L x 5 x 10^6 L = 1.5 x 10^9 mg = 1500 kg. (Key point: Mass of BOD)

## Close variants set in other exams

- [5-day BOD and kg BOD in 3.2 ML](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-dilution-2p5-in-250-7p8) — Sanitary Engineering, chapter 5, asked 2×

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- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
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