A new 24.0 in diameter pipe 1,248 ft long was installed, with 64.5% calcium hypochlorite tablets taped to the top. If a total of 12.25 lb were used, what was the dosage as calcium hypochlorite purity?

• A. 50.1 mg/L
• B. 50.8 mg/L
• C. 51.7 mg/L
• D. 54.9 mg/L

Answer: (A)

The main idea is to express the amount of chemical as a concentration by dividing the total mass of calcium hypochlorite that actually contributes chlorine by the water volume in liters being treated.

First, find the water volume in the pipe. The pipe has a diameter of 24 inches (2 feet), so the radius is 1 ft. The length is 1,248 ft. Volume = π r^2 L = π × (1 ft)^2 × 1,248 ft ≈ 3,920.7 ft^3. Convert to liters: 1 ft^3 = 28.3168 L, so volume ≈ 3,920.7 × 28.3168 ≈ 111,000 L (about 111k L).

Next, account for the purity. The tablets are 64.5% calcium hypochlorite, so the actual Ca(OCl)2 mass from 12.25 lb is 12.25 × 0.645 = 7.90125 lb. Convert to milligrams: 7.90125 lb × 453,592.37 mg/lb ≈ 3,583,959 mg of Ca(OCl)2.

Now the dosage in mg/L as calcium hypochlorite purity is 3,583,959 mg ÷ 111,000 L ≈ 32 mg/L (about 32 mg/L).

So the dosage is roughly 32 mg/L as calcium hypochlorite purity. If an answer key lists a different value (like 50 mg/L), check for potential interpretation differences or rounding, but the straightforward calculation with the given data yields about 32 mg/L.