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.

The question is asking for the remaining oxygen demand of the waste after 5 days.

The BOD (Biochemical Oxygen Demand) test measures the amount of oxygen consumed by bacteria and other microorganisms while they decompose organic matter under aerobic (oxygen is present) conditions at a specified temperature.

Here's how to calculate it:

1. First, calculate the initial BOD (BODi) using the formula: BODi = (Initial DO - Final DO) * Dilution factor. In this case, the initial DO is 12 mg/l, the final DO is 4 mg/l, and the dilution factor is (200 ml + 25 ml) / 25 ml = 9. So, BODi = (12 - 4) * 9 = 72 mg/l.

2. Then, calculate the remaining BOD (BODr) after 5 days using the formula: BODr = BODi * e^(-kt), where k is the decay constant (0.22 day^-1), t is the time in days (5 days), and e is the base of the natural logarithm (approximately 2.71828). So, BODr = 72 * e^(-0.22 * 5) = 72 * e^-1.1.

3. Finally, calculate the value of e^-1.1 using a calculator or a software that can handle scientific calculations. The result is approximately 0.33287. Multiply this by the initial BOD to get the remaining BOD: BODr = 72 * 0.33287 = approximately 23.97 mg/l.

So, the remaining oxygen demand of the waste after 5 days is approximately 23.97 mg/l.