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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» Solution :COD value is 40 ppm. It means 106g of water sample require 40 g of oxygen to oxidise the organic matter in it. ` 500g " water " to (40 XX 500)/(10^6)=2 xx 10^(-2) g " of " O_2` 500 mL water sample requires ` 2 xx 10^(-2) g ` of `O_2` , to oxidise the organic matter present in it . `8g O_2 -= 49 g K_2Cr_2O_7` ` 2 xx 10^(-2) g " of " O_2 -= (49 xx 2 xx 10^-2)/(8) g " of " K_2Cr_2O_7` Amount of `K_2Cr_2O_7` REQUIRED to oxidise the organic matter present in the water sample is 0.1225 g . |
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