Question

Problem: Using the data obtained in the lab, calculate the amount, in pounds, of Aluminum Sulfate that will be required to treat a holding tank that contains 14,500 gallons of wastewater. Assume that the Aluminum Sulfate contains 18 water molecules.

The Volume of each group is 1 liter

Answer 1

solution:

As we know that for a drinking water turbidity of water should lie in the range of 1 to 10 NTU that is why we select an appropriate dose of aluminium sulphate for raw water.

=> by using above given data of lab

=> for 1 liter of water 8ml of aluminium sulphate with 2ml of sodium carbonate is required.(turbidity is 1.06 NTU)
{1 gallon = 8.34 pound =3.785 Kg =3.79 litre (US)}

=> Calculation:

As we know that for a drinking water turbidity of water should lie in the range of 1 to 10 NTU that is why we select an appropriate dose of aluminium sulphate for raw water. = > by using above given data of lab = > for 1 liter of water 8ml of aluminium sulphate with 2ml of sodium carbonate is required.(turbidity is 1.06 NTU) {1 gallon = 8.34 pound = 3.785 Kg = 3.79 litre (US)} = > Calculation: step(1) Because 1 gallon = 3.79 litre of water at 4 degree c {us} Therefore 14500 gallon = 14500 times 3.79 = 54,955 litre of water But as per given information rightarrow 1 litre of water required 8 ml Al_2 (SO_4)_3. 18H_2 O therefore 1 litre H_2 O required = 8 ml Al_2 (SO_4)_3.18H_2 O then 54,955 litre = 8 times 54955 = 439640 ml of Aluminium sulphate Because 1 litre = 1000 ml Therefore 1 ml = 10^-3 litre. Therefore 439640 ml = 439.64 litre step(3) again 1 litre = 1/3.79 gallon OR 439.64 litre = 439.64/3.79 gallon = 116 gallon.