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10 mL of water required 1.47 mg of $K _{2}Cr _{2}O _{7}$ (M. wt. = 294) for oxidation of dissolved organic matter in presence of acid. Then the C.O.D of water sample is:

chemistry water too little to waste pollution of water water pollution planet earth

  1. 2.44 ppm

  2. 24 ppm

  3. 32 ppm

  4. 1.6 ppm

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Correct Option: B
Explanation:

$ Number \  of   \ milliequivalent $ = $\dfrac{weight \ in \ mg}{Equivalent\  weight}$


= $\dfrac{1.47}{294/6}$ = $0.03meq$

$\therefore$COD of sample is:

$\dfrac{0.03}{10}\times 1000$ =  $3meq/L$

In terms of Oxygen COD is :

$3 meq/L \times  (8 mg O _2/meq)$ = $24 mg/L= 24ppm$

Option B is correct.