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.14.6 mg/lit.35.9 mg/lt.68.2 mg/lit29.3 mg/lit.

The question seems to be asking for the remaining oxygen demand of the waste after 5 days. However, the question is not clear and seems to be missing some information.

In a BOD (Biochemical Oxygen Demand) test, the remaining oxygen demand can be calculated using the formula:

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

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

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

This is the oxygen demand after 5 days. However, this doesn't match any of the options provided in the question. It seems like there might be some missing or incorrect information in the question.