Please show me how to solve During drinking water treatment, 17 lb of chlorine a

Question

Please show me how to solve

During drinking water treatment, 17 lb of chlorine are added daily to disinfect 5 million gallons of watera) What is the aqueous concentration of chlorine in mM?

b) The chlorine demand is the concentration of chlorine used during disinfection. The chlorine residual is the concentration of chlorine that remains after treatment so that water maintains its disinfecting power in the distribution system. If the chlorine residual is 0.20 mg/L, what is the chlorine demand in mg/L?

Explanation / Answer

a) 1 lb = 453.592 g

   17 lb = 453.592 x 17 = 7711.064 g

Cl2 molar mass = 71 g / mol

Cl2 moles = mass / molar mass

                 = 7711.064 / 71.0

                 = 109

1 gallon = 3.78541 L

5 x 10^6 gallons = 5 x 10^6 x 3.78541 = 1.89 x 10^7 L

concentration = moles / volume

                       = 109 / 1.89 x 10^7

                       = 5.77 x 10^-6 M

concentration = 5.77 x 10^-6 M

1 M = 1000mM           

concentration = 5.77 x 10^-6 x 10^3 mM

concentration = 5.77 x 10^-3 mM      

concentration of Chlorine = 0.00577mM

(b)

above concentration = 0.00577 mM

just convert it into mg/L . to this multiply with 71 Cl2 molar mass

0.00577 mM   = 0.00577 mmol / L x (71 g /mol) = 0.40 mg / L

chlorine residual = 0.20 mg/L

chlorine demand = concentration Cl2 in water - residual chlorien

                              = 0.40 - 0.20

chlorine demand = 0.20 mg /L

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