Answer:
Water needs to be added into the pool at the rate of 1064.814 Liters/min.
Step-by-step explanation:
The concentration in mg/L of the chlorine in the pool that is induced by dumping the whole container of chlorine into the pool equals

Since this is in excess to the required concentration of 0.20 mg/L
Let the amount of pure water we need to add be equal to 'V' liters now since pure water does not have any chlorine thus
By the conservation of mass principle we have

Solving for V we get

Thus 2300000 Liters of water needs to be further added in 36 hours
Thus the rate of flow in L/min equals
