Determine the BOD5 for a 10 mL sample, collected from wastewater stream in Jamdoli that is diluted to a total volume of 300 mL when the initial dissolved oxygen is 8.5 mg/L and after 5 days, has been reduced to 2.5 mg/L.a.180 mg/Lb.140 mg/Lc.120 mg/Ld.160mg/L
Question
Determine the BOD5 for a 10 mL sample, collected from wastewater stream in Jamdoli that is diluted to a total volume of 300 mL when the initial dissolved oxygen is 8.5 mg/L and after 5 days, has been reduced to 2.5 mg/L.a.180 mg/Lb.140 mg/Lc.120 mg/Ld.160mg/L
Solution
The BOD5 (Biochemical Oxygen Demand over 5 days) is a measure of 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 formula to calculate BOD5 is:
BOD5 = (D1 - D2) * (Vt/Vs)
Where: D1 = Initial Dissolved Oxygen (DO) in mg/L D2 = Final DO after 5 days in mg/L Vt = Total volume of the diluted sample in mL Vs = Volume of the sample in mL
Given in the problem: D1 = 8.5 mg/L D2 = 2.5 mg/L Vt = 300 mL Vs = 10 mL
Substitute these values into the formula:
BOD5 = (8.5 mg/L - 2.5 mg/L) * (300 mL / 10 mL) BOD5 = 6 mg/L * 30 BOD5 = 180 mg/L
So, the BOD5 for the given sample is 180 mg/L. Therefore, the correct answer is (a) 180 mg/L.
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