Explanation:
The given data is as follows.
Volume of lake = =
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake =
= mg
= kg
Flow rate of river is 50
Volume of water in 1 day =
= liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are or
Flow rate of sewage =
Volume of sewage water in 1 day = liter
Concentration of sewage = 300 mg/L
Total amount of pollutants = or
Therefore, total concentration of lake after 1 day =
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence, =
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
