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.

Explanation

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 (40xx 500)/(10^4) = 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.
`2 xx 10^(-2) g of O_(2) -= (48 cc 2 xx 10^(-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.