Question

10 mL of water required 1.47 mg of $K_{2}C{r}_2{O}_7$ (M. wt. = 294) for oxidation of dissolved organic matter in presence of acid. Then the C.O.D of water sample is:

• A. 2.44 ppm
• B. 24 ppm
• C. 32 ppm
• D. 1.6 ppm

Answer 1

The correct option is B 24 ppm

$Numberofmilliequivalent$ = $\frac{weightinmg}{Equivalentweight}$

= $\frac{1.47}{294/6}$ = $0.03meq$
$\therefore$COD of sample is:

$\frac{0.03}{10}\times 1000$ = $3meq/L$
In terms of Oxygen COD is :

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

Option B is correct.