100 ml of the sample of water requires 3.92 mg of K₂Cr₂O₇ in presence of H₂SO₄ or the oxidation of dissolved organic matter present in it. The COD of the water sample in ppm

The milliequivalent of $K_2{Cr}_2{O}_7=\frac{3.92}{294}\times 6$

As $COD$ is reported in terms of equivalent oxygen, the equivalent weight of oxygen is 8.

Now,

By the law of equivalence.

$M$. eq. of $K_2{Cr}_2{O}_7=M$. eq. of $O_2$

Weight of $O_2=\frac{3.92\times 6}{294}\times \frac{32}{4}\times {10}^{-3}$

                        $$\begin{gathered} =0.64\times {10}^{-3}mg \\ =0.64 mg per100 mL \end{gathered}$$

$COD$ is the $mg$ of oxygen per liter sample of water.

Thus, $COD=6.4 mg per mL=6.4 ppm$