To determine the BOD5 of a waste water sample 10- 50- 75 and 100 ml samples of the waste water were diluted to 500 ml and incubated at 20-C in BOD bottles for 5 days- The results were as follows- S-No Waste water volume- ml Inital DO- mg-l Do after 5 days- mg-l- 1 10 9-8 6-9- 2 50 8-6 4-5- 3- 75 8-2 1-2- 4- 100 8-0 0-0- -Based on the above data- the average BOD5 of the waste water sample is mg-l

BOD in mg/l = [Initial Do Final DO] × Dilution factor For sample 1, [BOD_5]_1 = (9.8 6.9) ×50010 = 145 mg/l For sample 2, [BOD_5]_2 = (8.6 4.5) ×50050 = 4 ...

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To determine the $B O D_{5}$ of a waste water sample 10, 50, 75 and 100 ml samples of the waste water were diluted to 500 ml and incubated at $20^{\circ} C$ in BOD bottles for 5 days. The results were as follows:
$\begin{matrix} S.No & Waste water volume, ml & Inital DO, mg/l & Do after 5 days, mg/l 1 & 10 & 9.8 & 6.9 2 & 50 & 8.6 & 4.5 3. & 75 & 8.2 & 1.2 4. & 100 & 8.0 & 0.0 \end{matrix}$
Based on the above data, the average $B O D_{5}$ of the waste water sample is mg/l

77.56

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Solution

The correct option is A 77.56
BOD in mg/l = [Initial Do - Final DO] $\times$ Dilution factor
For sample 1, $[B O D_{5}]_{1} = (9.8 - 6.9) \times \frac{500}{10} = 145 m g / l$
For sample 2, $[B O D_{5}]_{2} = (8.6 - 4.5) \times \frac{500}{50} = 41 m g / l$
For sample 3, $[B O D_{5}]_{3} = (8.2 - 1.2) \times \frac{500}{75} = 46.67 m g / l$
Sample 4 not considered as its $D . O_{f i n a l} = 0$ and so, it is uncertain whether D.O. was used up in 5 days or 2 days or 3 days.
$∴$ Average $B O D_{5} = \frac{145 + 41 + 46.67}{3} = 77.56 m g / l$

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