Determine the carbonaceous and nitrogenous oxygen demand in mg/l for a 1 l solution containing 300 mg of a wastewater represented by the formula cn2h602 (n is converted to nh3 in the first step) afterwards calculate the cod of the solution

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

beta decay is the answer