Question

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

Answer 1

Answer:

COD  value is 40 ppm. It means 106g of water sample require 40g of oxygen to oxidise the organic matter in it.

500 water →40×500106=2×10−2g of O2

500 mL water sample requires 2×10−2g of O2 to oxidise the organic matter present in it.

2×10−2g of O2≡49×2×10−28g of K2Cr2O7

Amount of K2Cr2O7 required to oxidise the organic matter present in the water sample is 0.1225g.

Explanation: