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The `COD` value of a water sample is 40 ppm. Calculate the amount of acidified `K_(2) Cr_(2) O_(7)` required to oxidise the organic matter present in 500 ml of that water sample. |
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Answer» `COD` value is 40 ppm. It means `10^(6) g` of water sample require `40 g` of oxygen to oxidise the organic matter in it. `500` water `rarr (40xx500)/10^(6)=2xx10^(-2) g` of `O_(2)` `500` mL water sample requires `2xx10^(-2) g` of `O_(2)` to oxidise the organic matter present in it. `2xx10^(-2) g` of `O_(2) equiv (49xx2xx10^(-2))/(8) g` of `K_(2)Cr_(2)O_(7)` Amount of `K_(2) Cr_(2) O_(7)` required to oxidise the organic matter present in the water sample is `0.1225 g`. |
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