Question

A sample of municipal sewage is diluted to 1% by volume prior to running a BOD5 analysis. After 5 days the oxygen consumption is determined to be 2.00 mg · L−1. What is the BOD5 of the sewage

Answer 1

Answer:

BOD5 = 200 mg/L

Explanation:

given data

diluted = 1% = 0.01

time = 5 day

oxygen consumption = 2.00 mg · L−1

solution

we get here BOD5  that is BOD after 5 day

and here total volume is 100% = 1

so dilution factor is $\frac{100}{1}$    =  100

so BOD5 is

BOD5 = oxygen consumption × dilution factor

BOD5 = 2 × 100

BOD5 = 200 mg/L