Question

If 2% solution of a sewage sample is incubated for 5 days at 20°C and depletion of oxygen was found to be 5 ppm, then the B.O.D. of sewage is

• A. 200 ppm
• B. 225 ppm
• C. 250 ppm
• D. 275 ppm

Hint:

B.O.D. calculations involving dilution require the dilution factor, which is the ratio of the total volume of the diluted sample to the volume of the sewage. For a 2% solution, the dilution factor is 50, meaning the B.O.D. of the undiluted sewage is 50 times the measured B.O.D. of the diluted sample.

Answer 1

The Correct Option is C

Step 1: Identify the given data.
A 2% solution of a sewage sample is incubated for 5 days at 20°C, and the oxygen depletion (B.O.D. of the diluted sample) is 5 ppm. We need to find the B.O.D. of the sewage (undiluted sample) in ppm. The options are:
(1) 200 ppm
(2) 225 ppm
(3) 250 ppm
(4) 275 ppm
Step 2: Recall the concept of B.O.D. and dilution.
The Biochemical Oxygen Demand (B.O.D.) measures the amount of oxygen consumed by microorganisms to decompose organic matter in the sewage over a specific period (typically 5 days at 20°C, denoted as \(BOD_{5}\)). When a sewage sample is diluted, the measured B.O.D. of the diluted sample must be adjusted to find the B.O.D. of the undiluted sewage using the dilution factor.
A 2% solution means 2 parts of sewage are mixed with 98 parts of water, making a total of 100 parts. The dilution factor (\( D \)) is:
\[ D = \frac{\text{Total volume}}{\text{Volume of sewage}} = \frac{100}{2} = 50 \] The B.O.D. of the undiluted sewage is:
\[ \text{B.O.D.}_{\text{sewage}} = \text{B.O.D.}_{\text{diluted}} \times \text{Dilution factor} \]
Step 3: Calculate the B.O.D. of the sewage.
The B.O.D. of the diluted sample (2% solution) is given as 5 ppm. Using the dilution factor:
\[ \text{B.O.D.}_{\text{sewage}} = 5 \, \text{ppm} \times 50 = 250 \, \text{ppm} \]
Step 4: Verify the calculation.
The 2% solution means the sewage is diluted 50 times (since \( \frac{2}{100} = 0.02 \), and the dilution factor is \( \frac{1}{0.02} = 50 \)). The oxygen depletion of 5 ppm in the diluted sample scales up by the dilution factor to give the B.O.D. of the original sewage, confirming 250 ppm.
Step 5: Select the correct option.
The B.O.D. of the sewage is 250 ppm, which matches option (3).
\[ \boxed{250 \, \text{ppm}} \]