An effluent obtained from an industry has total BOD of 120mg/l. A sample of 50ml from effluent is taken to run BOD test. The size of bottle is 300ml. The final and initial dissolved oxygen of the sample are $9 \frac{m g}{l}$ and $21 \frac{m g}{l}$ . The nitrogenous BOD (in $\frac{m g}{l}$ of the effluent is :

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Solution

The correct option is B 48
Carbonaceous BOD of sample= $(D O_{f} - D O_{i}) \times D i l u t i o n f a c t o r$
$= (21 - 9) \times \frac{300}{50}$
$= 72 \frac{m g}{l}$
$N - B O D = T o t a l B O D - C B O D$
$= (120 - 72) \frac{m g}{L}$
$= 48 \frac{m g}{l}$

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An effluent obtained from an industry has total BOD of 120mg/l. A sample of 50ml from effluent is taken to run BOD test. The size of bottle is 300ml. The final and initial dissolved oxygen of the sample are 9mgl and 21mgl. The nitrogenous BOD (in mgl of the effluent is :

The correct option is B 48Carbonaceous BOD of sample= (DOf−DOi)×Dilutionfactor =(21−9)×30050 =72mgl N−BOD=Total BOD−C BOD =(120−72)mgL =48mgl

The correct option is B 48
Carbonaceous BOD of sample= $(D O_{f} - D O_{i}) \times D i l u t i o n f a c t o r$
$= (21 - 9) \times \frac{300}{50}$
$= 72 \frac{m g}{l}$
$N - B O D = T o t a l B O D - C B O D$
$= (120 - 72) \frac{m g}{L}$
$= 48 \frac{m g}{l}$