Question

$10$mL of water required $1.47$mg of $K_{2}C{r}_2{O}_7$ (mol. wt. $=294$) for oxidation of dissolved organic matter. C.O.D (Chemical Oxygen Demand) of water is?

• A. $3.44$ ppm
• B. $24$ ppm
• C. $34.4$ ppm
• D. $2.44$ ppm

Answer 1

The correct option is B $24$ ppm

$10$mL water $=1.47\times {10}^{-3}$g of $K_{2}C{r}_2{O}_7$
${10}^6$mL water $=1.47$g of $K_{2}C{r}_2{O}_7$
$\because 49$g of $K_{2}C{r}_2{O}_7$ requires $O_2=89$
$\therefore 147$g of $K_{2}C{r}_2{O}_7$ will required $O_2=\frac{8\times 147}{49}$
$=24$g.