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.29.3 mg/lit.68.2 mg/lit

The question is 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, which is the total demand minus the amount of oxygen already consumed. The amount of oxygen consumed is the initial DO minus the final DO, which is 12 mg/l - 4 mg/l = 8 mg/l.

So, the remaining oxygen demand is 72 mg/l - 8 mg/l = 64 mg/l.

However, none of the given options match this result. There might be a mistake in the question or the given options.