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 is ans. 29.3 mg/lit. 35.9 mg/lt. 14.6 mg/lit. 68.2 mg/lit

The Biochemical Oxygen Demand (BOD) 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.

The BOD equation is:

BOD = (DOi - DOf) * (Vt/Vs)

Where: DOi = Initial Dissolved Oxygen DOf = Final Dissolved Oxygen Vt = Total volume of the sample Vs = Volume of the sample of waste water

Given: DOi = 12 mg/l DOf = 4 mg/l Vt = 225 ml (200 ml of pure water + 25 ml of waste water) Vs = 25 ml

Substituting the given values into the BOD equation:

BOD = (12 mg/l - 4 mg/l) * (225 ml/25 ml) BOD = 8 mg/l * 9 BOD = 72 mg/l

The remaining oxygen demand after 5 days is the initial BOD minus the oxygen consumed after 5 days.

The oxygen consumed after 5 days can be calculated using the first order reaction equation:

BOD_remaining = BOD_initial * e^(-kt)

Where: k = deoxygenation rate constant = 0.22 day^-1 t = time = 5 days

Substituting the given values into the equation:

BOD_remaining = 72 mg/l * e^(-0.22 day^-1 * 5 days) BOD_remaining = 72 mg/l * e^-1.1 BOD_remaining = 72 mg/l * 0.33287 BOD_remaining = 23.97 mg/l

Therefore, the remaining oxygen demand of the waste after 5 days is approximately 24 mg/l. However, this value is not listed in the provided options. There might be a mistake in the question or the provided options.