If the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C, B.O.D. of the sewage is A. 50 mg/l B. 100 mg/l C. 150 mg/l D. 250 mg/l

50 mg/l
100 mg/l
150 mg/l
of the sewage is A. 50 mg/l B. 100 mg/l C. 150 mg/l D. 250 mg/l

The correct answer is A. 50 mg/l.

Biological oxygen demand (BOD) is the amount of oxygen required by microorganisms to decompose organic matter in a given water sample at a specified temperature over a specified time period. It is a measure of the organic pollution in water.

The BOD test is conducted by incubating a sample of water with a known amount of dissolved oxygen for a specified time period, usually five days at 20°C. The amount of oxygen consumed during the incubation period is then measured.

The BOD of a sample of water is expressed in milligrams of oxygen per liter (mg/l). A high BOD indicates that the water is polluted with organic matter.

In the question, the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C. The BOD of the sewage is therefore 2.5 x 100 = 50 mg/l.

Option A is the correct answer because it is the only option that is within the range of BOD values for sewage. The BOD of sewage typically ranges from 100 to 500 mg/l.

Option B is incorrect because it is twice the actual BOD value.

Option C is incorrect because it is three times the actual BOD value.

Option D is incorrect because it is five times the actual BOD value.