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 oxidies the organic matter in it .
500g water `rarr (40xx500)/(10^6)=2xx10^(-2)g " of " O_2`
500 mL water sample requires `2xx10^(-2)" g of " O_(2)` , to oxidies the organic matter present in it.
`8gO_(2)-= 49 g K_(2)Cr_(2)O_(7)`
`2xx10^(-2)g of O_2-=(49xx2xx10^(-2))/8g " of " K_(2)Cr_(2)O_(7)`
Amount of `K_(2)Cr_(2)O_(7)` required to oxidies the organic matter present in water sample is 0.1225 g .