COD of a water sample is 8ppm the weight of acidified $K_{2}C{r}_2{O}_7$ required to oxidize the organic matter present in one litre of water sample is______ mg.

COD of water sample $=8ppm$

${10}^{6}g$ of $H_{2}O=8g$ of $O_2$

So, $1L=\frac{8\times {10}^3}{{10}^6}=8\times {10}^{-3}g$ of $O_2$

Now, 8-gram equivalence of $O_2=49$ gram equivalence of $K_2{Cr}_2{O}_7$

For, $8\times {10}^{-3}$ of $O_2=\frac{49\times 8\times {10}^3}{8}=49\times {10}^{-3}=49mg$

Hence, for $1L$ of water sample, the required weight is $49mg$.