Question

A 104 m3 thoroughly mixed pond has a water inflow and outflow of 5 m3/h. The inflow water contains 0.01 mol/m3 of chemical. Chemical is also discharged directly into the pond at a rate of 0.1 mol/h. There is no reaction, volatilization, or other losses of the chemical; it all leaves in the outflow water. (i) What is the concentration (C) in the outflow water?

Answer 1

C ≈ 1.44 × 10⁻³ mol/m³

Step-by-step explanation:

The given information are;

The liquid volume the pond can hold = 104 m³

The volume of inflow into the pond = 5 m³/h

The volume of outflow into the pond = 5 m³/h

The concentration of the chemical in the inflow water = 0.01 mol/m³

The concentration of the chemical discharged directly into the water = 0.1 mol/h

The concentration,
`c_((inflow))`, of chemical that enters the water through inflow per hour is given as follows;

`c_((inflow))` = 0.01 mol/m³ × 5 m³/h = 0.05 mol/h

The concentration,
`c_((discharge))`, of chemical that enters the water through direct discharge per hour is given as follows;

`c_((discharge))` = 0.1 mol/h

The total concentration that enters the pond per hour is given as follows;

`c_((inflow))` +
`c_((discharge))` = 0.1 mol/h + 0.05 mol/h = 0.15 mol/h

Whereby the water in the pond properly mixes with the pond, we have;

The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³

C ≈ 1.44 × 10⁻³ mol/m³.