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