`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`.