Q80 / 180 · JIPMER 2015 · Chemistry

500 mL of a sample of water required 19.6 mg of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ for the oxidation of dissolved organic matter in it in the presence of $\mathrm{H}_2 \mathrm{SO}_4$. The COD of water sample is

  • A
    3.2 ppm
  • B
    7.2 ppm
  • Correct answer
    C
    6.4 ppm
  • D
    4.6 ppm

Solution

$\begin{aligned} & 500 \mathrm{~mL} \mathrm{H} \mathrm{H}_2 \mathrm{O} \longrightarrow 19.6 \times 10^{-3} \mathrm{~g} \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \\ & 106 \mathrm{~mL} \mathrm{H}_2 \mathrm{O} \longrightarrow 39.2 \mathrm{~g} \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \\ & 49 \mathrm{~g} \text { of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \longrightarrow 8 \mathrm{~g} \text { of } \mathrm{O}_2 \\ & 39.2 \mathrm{~g} \text { of } \mathrm{K}_2 \mathrm{C}_2 \mathrm{O}_7 \longrightarrow 6.4 \mathrm{~g} \text { of } \mathrm{O}_2 \\ & \therefore \quad \text { COD }=6.4 \mathrm{ppm} \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.

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