Question

In a biochemical oxygen demand (BOD) test, 15 ml of wastewater sample was diluted with distilled water to completely fill a 300 ml BOD bottle and incubated at 20 \(^{\circ}\text{C}\) for 5 days. The dissolved oxygen (DO) level before and after incubation are 9.2 mg liter\(^{-1}\) and 4.4 mg liter\(^{-1}\), respectively. The BOD of the sample, in mg liter\(^{-1}\), is . (answer in integer)

Hint:

Scale the observed drop in dissolved oxygen up by the dilution factor of the BOD bottle to get the BOD of the original sample.

Answer 1

Correct Answer: 96

Step 1: Understanding the Concept:
The sample was too concentrated to test directly, so a small part of it, 15 ml, was diluted with clean water to fill the whole 300 ml bottle. This means the sample only makes up a fraction of the bottle's contents.

Step 2: Key Formula or Approach:
The fraction of the bottle that is actual sample is $f = 15/300 = 1/20$. The oxygen drop measured in the bottle was caused only by this fraction of sample, so the BOD of the full strength sample is the measured drop divided by that fraction: $BOD = \Delta DO / f$.

Step 3: Detailed Explanation:
The dissolved oxygen fell from 9.2 mg/L before incubation to 4.4 mg/L after 5 days, a drop of $\Delta DO = 9.2 - 4.4 = 4.8$ mg/L.
The sample fraction is $f = 15/300 = 0.05$.
Dividing the observed drop by that fraction, $BOD = 4.8 / 0.05 = 96$ mg/L.
This matches scaling up by the reciprocal of the fraction, $1/0.05 = 20$, which is the same dilution factor used the other way around.

Step 4: Final Answer:
The five day BOD of the wastewater sample works out to 96 mg per liter.