Answer to Question #164921 in General Chemistry for Lisa Middlebrook

Question

A home owner has a pond and uses a mixture that contains 11.3% Bromine by mass to kill an

algae bloom. An ideal Bromine level for the pond is 3 parts per million (3 ppm). (Think of 1

ppm as being 1 g Bromine per million grams of water.) If you assume densities of 1.17 g/mL

for the Bromine solution and 1.00 g/mL for the pond water, what volume of the bromine

solution, in liters, is required to produce a bromine level of 3 ppm in a 50,000-gallon pond

Answer 1

50000 gallons = 189× 10⁶ ml water = 189×10⁶ g water.

Since it produces 3 g bromine in one million grams

So total bromine = 189×3 = 567 g

Volume of bromine = 567/1.17 mL = 484.6 mL