A sample of pond water containing 40mg of organic matter requires 32mg of dissolved oxygen. If pond water contains 100 mg of organic matter per two liters, the BOD value of the water sample is______ ppm.

100mg of organic matter contains = 2L water

So, $40mg$ of organic matter contains $=\frac{2\times 40}{100}=0.8L$

Now, we know,

$BOD=\frac{No.ofmilligramsofrequiredoxygen}{volumeofwater}=>BOD=\frac{32mg}{0.8 L}=40ppm$

Thus, the BOD value of the water sample is $40 ppm$.