Question

Calcium hypochlorite (ca(ocl)2, mw = 142.983 g/mol) is often used as the source of the hypochlorite ion (ocl–, mw = 51.452 g/mol) in solutions used for water treatment. a student must prepare 50.0 ml of a 40.0 ppm ocl– solution from solid ca(ocl)2, which has a purity of 95.0%.

Answer 1

When the concentration is expressed in ppm, that means parts per million. It is also equivalent to mg/L. For this problem, we do stoichiometric calculations. We manipulate the units by cancelling like units if they appear in the numerator and denominator side until we come with the amount of solid Ca(OCl)2 needed. The solution is as follows:

40 mg/L * (1 L/1000 mL) * 50 mL * (1 g/1000 mg) * (1 mol OCl⁻/51.452 g) * (1 mol Ca(OCl)₂/ 2 mol OCl⁻) * (142.983 g Ca(OCl)₂/mol) * 0.95 = 2.64×10⁻3 g or 2.64 mg.

Therefore, you would need 2.64 mg of solid Ca(OCl)₂.