$$100$$ ml of sample of water requires $$3.92$$ mg of $$K2Cr2$$ $$O7$$ in presence of $$H2SO4$$ for the oxidation of dissolved organic matter present in it. The COD of the water sample in ppm

$$COD=COD=wt$$ .of $$K2Cr2O7$$×$$8$$×$$10649$$×sample of water in $$ml3.92$$×$$10-3$$×$$8$$×$$10649$$×$$100=6.4$$

Water pollution

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Question

100 ml of sample of water requires 3.92 mg of K₂Cr₂ O₇ in presence of H₂SO₄ for the oxidation of dissolved organic matter present in it. The COD of the water sample in ppm

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Solution

COD= $COD = \frac{wt . of K_{2} {Cr}_{2} O_{7} \times 8 \times 10^{6}}{49 \times sample of water (in ml)} \frac{3.92 \times 10^{- 3} \times 8 \times 10^{6}}{49 \times 100} = 6.4$

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