---
title: "5-day BOD and kg BOD in 3.2 ML"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-dilution-2p5-in-250-7p8"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-dilution-2p5-in-250-7p8
site: BCT Engineering
---

# 5-day BOD and kg BOD in 3.2 ML

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2070 Bhadra, 8 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-250-7p8
- 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. 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.

## 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.

## Answer

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

#### 5-day BOD of raw sewage
$$
BOD_5=\frac{D_1-D_2}{P}=\frac{7.8-5.1}{0.01}=270\ \text{mg/L}
$$

#### Mass of BOD in 3.2 million litres
$$
\text{Mass}=270\ \text{mg/L}\times3.2\times10^{6}\ \text{L}=864000000\ \text{mg}=864\ \text{kg}
$$

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

## Check yourself

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

1. Why is the scanned value 51.0 read as 5.1 mg/L?
   A. Final DO cannot exceed the initial 7.8 mg/L
   B. It is more than the blank
   C. It is below zero
   D. BOD must be 51
   Answer: A. DO falls during incubation, so 51.0 is impossible. (Key point: Reading check)

2. What is P?
   A. 250/2.5 = 100
   B. 2.5/250 = 0.01
   C. 0.1
   D. 0.0025
   Answer: B. P is the sample fraction in the mixture. (Key point: Dilution fraction)

3. What is BOD5?
   A. 2.7 mg/L
   B. (7.8 + 5.1)/0.01
   C. (7.8 - 5.1)/0.01 = 270 mg/L
   D. 7.8/0.01 = 780 mg/L
   Answer: C. DO drop divided by P. (Key point: BOD5)

## Close variants set in other exams

- [5-day BOD and kg BOD in 5 ML](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-dilution-2p5-in-250-8mg) — Sanitary Engineering, chapter 5, asked 2×

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