0.5 ppm = 0.5 g Cl2 in 1,000,000 g solution. The density of water is essentially 1 g/mL and the mass of Cl2 is so small that 1 million grams of water + small amount Cl2 = 1 million grams solution or 1 million mL.
So we can think in terms of 0.5 g Cl2/10^6 mL water.
Convert that to grams Cl2/L. (that's 0.5 g Cl2/1000 mL), then to 100 L.
What mass of cholrine must be added to 100.0 L of water to achieve this level? Most community water supplies have 0.5 ppm of cholrine added for purification.
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