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500 mL of a sample of water required 19....

500 mL of a sample of water required 19.6 mg of `K_2Cr_2O_7` for the oxidation of dissolved organic matter in it in the presence of `H_2SO_4`. The COD of water sample is

A

8 ppm

B

6.4 ppm

C

16.8 ppm

D

4.9 ppm

Text Solution

Verified by Experts

The correct Answer is:
B
500 mL `H_2O - 19.6xx10^(-3) g K_2Cr_2O_7`
`10^6` mL `H_2O` - 39.2 g `K_2Cr_2O_7`
49 g of `K_2Cr_2O_7` - 8 g of `O_2`
39.2 g of `K_2Cr_2O_7` - 6.4 g of `O_2`
`therefore` COD=6.4 ppm
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