$100 m l$ of a sample of water requires $0.98 m g$ of $K_{2} {C r}_{2} O_{7}$ (M.W. $= 294$ ) in presence of the $H_{2} {S O}_{4}$ for the oxidation of dissolved organic matter in it. The C.O.D of the water sample (in $p p m$ ) is?

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$78.4$

$1.6$

$3.2$

$6.4$

answer is B.

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Detailed Solution

Following is the reaction that will be taking place; $K_{2} {C r}_{2} O_{7} + 4 H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} ({S O}_{4})_{3} + 4 H_{2} O + 3 (O)$

Now,

$294 g$ of $K_{2} {C r}_{2} O_{7}$ requires $3 \times 16 g$ of $(O) = 48 g$ of $(O)$

Now,

$1 g$ of $K_{2} {C r}_{2} O_{7}$ will require $0.163 g$ of $(O)$

Thus,

$0.98 m g$ will require $0.16 m g$ of $(O)$ .

This $0.16 m g$ of $(0)$ is for a $100 m l$ sample.

Now, we know that $C O D$ is defined as the number of oxygen equivalents consumed in the chemical oxidation of organic matter by a strong oxidant (e.g. potassium dichromate).

So,

$C . O . D = 0.16 \times \frac{1000}{100} p p m = 1.6 p p m$