An unseeded BOD test is run using 25 ml of waste water mixed with 200 ml of pure water. The value of k at 20℃ is 0.22 day-1. The initial DO of the mixture is 12 mg/l. After 5 days the DO remains fixed at 4 mg/l. The remaining oxygen demand of the waste after 5 days isans.35.9 mg/lt.14.6 mg/lit.68.2 mg/lit29.3 mg/lit. Previous Marked for Review Next

The question seems to be asking for the remaining oxygen demand of the waste after 5 days. This can be calculated using the formula for Biological Oxygen Demand (BOD):

BOD = (Initial DO - Final DO) * Dilution Factor

The initial DO (Dissolved Oxygen) is given as 12 mg/l, and the final DO after 5 days is 4 mg/l. The dilution factor is the total volume of the mixture divided by the volume of the waste water, which is (200 ml + 25 ml) / 25 ml = 9.

So, BOD = (12 mg/l - 4 mg/l) * 9 = 72 mg/l

However, this is the total oxygen demand after 5 days. The question asks for the remaining oxygen demand. This can be calculated using the formula:

Remaining BOD = Total BOD * e^(-kt)

where k is the decay constant (0.22 day^-1) and t is the time (5 days).

So, Remaining BOD = 72 mg/l * e^(-0.22 day^-1 * 5 days) = 35.9 mg/l

Therefore, the remaining oxygen demand of the waste after 5 days is approximately 35.9 mg/l.