I need your help please During drinking water treatment, 17 lb of chlorine are a

Question

I need your help please

During drinking water treatment, 17 lb of chlorine are added daily to disinfect 5 million gallons of water. a) 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

Given :

Mass of Cl2 = 17 lb

Volume of water = 5 milliion gallons

Solution :

We have to find concentration of Cl2 in mM

mM is for mili molar

We know concentration ( M ) = # mol / Volume in L

We get moles of Cl2 from its given mass and molar mass of Cl2

Lets convert 17 lb to g

1 lb = 453.592 g

Mass in g = 17 lb * 453.592 g / 1 lb = 7711.064 g

Molar mass of Cl2 = 70.906 g/mol

#mol Cl2 = Mass in g / Molar mass = 7711.064 g / 70.906 g per mol

= 108.751 mol

Volume of water in L

1 gallon = 3.7854 L

Lets convert volume from gallons to L

5 million gallons = 5 X 106 gallons

Volume in L

= 5 X 106 gallons * 3.7854 L / 1gallon

= 1.89 X 107 L

Now lets find concentration in M

Concentration = # mol / volume in L

= 108.751mol / 1.89 X 107 L

= 5.75 X 106 M

But we need concentration in mM

1 M = 1000 mM

Concentration in mM

= 5.75 X 106 M * 1000 mM/ 1 M

= 5.75 X 109 mM

Chlorine demand

=Concentration of chlorine (part a)

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