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 40 ppm means `10^(6)g` of water sample require 40 g of oxygen to oxidise the organic matter in it.  
`10^(6)g"water "rarr+" 40g "O_(2)" 500g water"rarr(40xx500)/(10^(6))=2xx10^(-2)"g of "O_(2)`
`"8g "O_(2)-="49 g "K_(2)Cr_(2)O_(7),2xx10^(-2)g O_(2)-=(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 0.1225g.