Calcium hypochlorite (Ca(OCl)_2 MW = 142.983 g/mol) is often used as the source
ID: 477742 • Letter: C
Question
Calcium hypochlorite (Ca(OCl)_2 MW = 142.983 g/mol) is often used as the source of the hypochlorite ion (OCI^- MW = 51.452 g/mol) in solutions used for water treatment. A student must prepare 50.0 mL of a 25.0 ppm OCI solution from solid Ca(OCI)_2, which has a purity of 93.0%. Calculate the mass of the impure reagent required to prepare the solution. Assume the density of the solution is 1.00 g/mL. Which of the following methods would work best to accurately prepare 50.0 ml_ of the 25.0 ppm OCI^- solution? Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 25.0 ppm. Use a pipet to transfer a portion of the concentrated solution lo a separate 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 25.0 ppm. Use an analytical balance to weigh out the amount of reagent determined in part A. Q Transfer the reagent to a 50.0 mL volumetric flask, add water to the line, and mix thoroughly. Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create O a solution with a concentration greater than 25.0 ppm. Use a graduated cylinder to transfer a portion of the more concentrated cylinder to transfer a portion of the more concentrated solution to a 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 25.0 ppm. Use an analytical balance to weigh out the amount of reagent determined in part A. 0 Then use a graduated cylinder to measure out 50.0 mL of water. Combine the reagent and water in a 50.0 mL volumetric flask and mix it thoroughly.Explanation / Answer
complete ionization
Ca(OCl)2 Ca{2+} + 2OCl{-}
(50.0 ml)* (1 g/moL) * (25.0 * 10^-6) / ( 51.452 g ocl{-}/mol) * ( 1 mol Ca(OCl)2 / 2 mol OCl{-} *
(142.983 g ca(OCl)2 / mol) / (0.930)
0.00125 / 0.1799 / 0.930 = 0.00747 impure Ca(OCl)2