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Q.
$10\, mL$ of water requires $1.47\,mg$ of $K_2Cr_2O_ 7$ (M. wt. $= 294$) for oxidation of dissolved organic matter. $C.O.D$ is
Environmental Chemistry
Solution:
$10 mL$ water $=1.47 \times 10^{-3} g$ of $K _{2} Cr _{2} O _{7}$
$10^{6} mL$ water $=1.47 g$ of $K _{2} Cr _{2} O _{7}$
$\because 49 g$ of $K _{2} Cr _{2} O _{7}$ requires $O _{2}=89$
$\therefore 147 g$ of $K _{2} Cr _{2} O _{7}$ will required
$O _{2}=\frac{8 \times 147}{49}$
$=24 g .$