Question

If the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250ml for 5days at 20°c



b.O.D of sewage is

Answer 1

Explanation:

Depletion of oxygen=5ppm

Solution of sewage=2%

Consider,

Total quantity of sewage= 100%

We have,

BOD = (DO initial -DO final) * Dilution factor.

Dilution factor =(Vol of diluted sample/Vol of in diluted sewage sample).

DO initial-DO final means depleted oxygen content.

Therefore,

BOD = 5 * (100/2) = 250ppm.