Question

If the depletion of oxygen is found to be 5ppm after incubating a 2.5% solution of sewage sample for 5 days at 21°c,



b.O.D of the sewage

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.