JEE Advanced levelNumerical answerDifficulty 3 of 3ev153
A 12.0mL portion of waste water is made up to 300mL with oxygen-saturated dilution water. The diluted mixture has dissolved oxygen of 8.7mgL−1 at the start and 4.9mgL−1 after incubation for 5 days at 20 ∘C. The BOD5 of the original waste water is calculated as (initial DO minus final DO) times the dilution factor. Give the BOD5 in mgL−1 to the nearest integer.
Numerical answer type, answer in mg/L, tolerance 0.5.
Answer: 95 mg/L
Dilution factor = 300/12.0=25. DO drop = 8.7−4.9=3.8 mg/L. BOD5=3.8×25=95 mg/L, which is far above 17 ppm, so the waste water is heavily polluted.