If a facility treats 5.5 million gallons of water with an average alum dosage of 7.34 mg/L, how many grams of alum are required daily?

If a facility treats 5.5 million gallons of water with an average alum dosage of 7.34 mg/L, how many grams of alum are required daily?

• A. 22.6
• B. 3.3
• C. 46.3
• D. 25.0

Answer: (C)

To determine the daily amount of alum required in grams when treating a facility with 5.5 million gallons of water and an average alum dosage of 7.34 mg/L, you need to follow these conversion steps. First, convert the volume of water from million gallons to liters, since the alum dosage is expressed in mg/L. One million gallons is approximately equal to 3,785,411.78 liters. Therefore, 5.5 million gallons is equivalent to: 5.5 million gallons × 3,785.41 L/million gallons = 20,817,253.5 liters. Next, multiply the volume of water by the dosage of alum (in mg/L). The alum dosage of 7.34 mg/L means that for every liter of treated water, 7.34 milligrams of alum are required. So, you calculate the total mass of alum needed by: Total alum needed (in mg) = Volume of water (in liters) × Alum dosage (in mg/L) Total alum needed (in mg) = 20,817,253.5 L × 7.34 mg/L = 152,353,388.3 mg. Now convert milligrams to grams since there are

To determine the daily amount of alum required in grams when treating a facility with 5.5 million gallons of water and an average alum dosage of 7.34 mg/L, you need to follow these conversion steps.

First, convert the volume of water from million gallons to liters, since the alum dosage is expressed in mg/L. One million gallons is approximately equal to 3,785,411.78 liters. Therefore, 5.5 million gallons is equivalent to:

5.5 million gallons × 3,785.41 L/million gallons = 20,817,253.5 liters.

Next, multiply the volume of water by the dosage of alum (in mg/L). The alum dosage of 7.34 mg/L means that for every liter of treated water, 7.34 milligrams of alum are required. So, you calculate the total mass of alum needed by:

Total alum needed (in mg) = Volume of water (in liters) × Alum dosage (in mg/L)

Total alum needed (in mg) = 20,817,253.5 L × 7.34 mg/L = 152,353,388.3 mg.

Now convert milligrams to grams since there are