The COD value of a water sample is 40ppm. Calculate the amount of acids fired K2Cr2O7 required to oxidised the organic matter present in 500ml of that water sample

106gH2O→40gO2500g water →40×500106=2×10−2g of O28gO2=49gK2Cr2O72×10−2gO2=49×2×10−28g of K2Cr2O7=0.1225g

The COD value of a water sample is 40ppm. Calculate the amount of acids fired K2Cr2O7 required to oxidised the organic matter present in 500ml of that water sample

106gH2O→40gO2500g water →40×500106=2×10−2g of O28gO2=49gK2Cr2O72×10−2gO2=49×2×10−28g of K2Cr2O7=0.1225g

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The COD value of a water sample is 40ppm. Calculate the amount of acids fired K₂Cr₂O₇ required to oxidised the organic matter present in 500ml of that water sample

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Detailed Solution

$\begin{array}{l} 10^{6} {gH}_{2} O \to 40 {gO}_{2} \\ 500 g water \to \frac{40 \times 500}{10^{6}} = 2 \times 10^{- 2} g of O_{2} \\ 8 {gO}_{2} = 49 {gK}_{2} {Cr}_{2} O_{7} \\ 2 \times 10^{- 2} {gO}_{2} = \frac{49 \times 2 \times 10^{- 2}}{8} g of \\ K_{2} {Cr}_{2} O_{7} = 0 .1225 g \end{array}$