**Explanation:**

It is given that in 1 liter of solution, there is 300 g of wastewater in .

Therefore, the reaction equation will be as follows.

For carbonaceous oxygen demand:

Molecular weight of =

= 174 g/mol

or, = mg/mol (as 1 g = 1000 mg)

So, total moles present in 300 mg of solution will be as follows.

Moles =

=

=

Hence, for 1 mol of carbonaceous oxygen demand is B mol.

Therefore, for oxygen required is calculated as follows.

= 0.0138 mol

Weight of oxygen required (m) =

= 0.4414 g

or, = 441.4 mg

**This means COD (carbonaceous oxygen demand) is 441.4 mg.**

For nitrogenous oxygen demand:

For 1 mol of to convert into , oxygen required is 2 mol.

For of , oxygen is calculated as follows.

= 0.006896 mol

Mass of required is 0.22 g = 220.6 mg

NOD (Nitrogenous oxygen demand) =

Therefore,** calculate total biological oxygen demand (BOD) as follows.**

=

= 662 mg /L