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## How do I prepare a solution with concentration in ppm from a solution with concentration in percent?
**I have 100 ml of Clorox containing 5.25% NaOCl and want to make a 1000 ml of total solution (d.i. water and clorox) containing 200 ppm of NaOCl??? How do I start???** Jasmine-
Solve this problem either by using dilution factors or by setting up a unit conversion calculation.**Vocabulary**dilution mass percentage parts per million solute solution solvent volume percentage
**The quick way (dilution factor method)**. Think of percentages as 'parts per hundred'. One part per hundred is 10,000 parts per million, so 5.25% NaOCl is 5.25×10^{4}ppm. Preparing a 200 ppm solution requires dilution by a factor of 200 ppm/ 5.25×10^{4}ppm or 1/263. That means that the mass of Chlorox that must be diluted to 1000 mL is 1/263 of 1000 g (the approximate mass of the diluted solution).**The mechanical way (unit conversion method)**. Take one step at a time. I'll assume the percent and ppm are in terms of masses, not volumes [1].**Pick out the target**. You must dilute some of the Chlorox by adding water to it. Rephrase the question: How much Chlorox (5.25% NaOCl) should be diluted to 1000 mL to make a 200 ppm NaOCl solution? Find*g of Chlorox*.? g of Chlorox **List the given information, and connect it with the target**. Try to work backwards from the target. You know two things about the Chlorox solution: you have 100 mL, and it's 5.25% NaOCl. The latter means, "5.25 g of NaOCl are found in 100 g of Chlorox". That connects the target with g of NaOCl.? g NaOCl 5.25 g NaOCl = 100 g Chloroxg Chlorox ? g solution 200 g NaOCl = 10^{6}g solutiong NaOCl 5.25 g NaOCl = 100 g Chloroxg Chlorox 1000 mL solution 1.00 g solution = 1 mL solutiong solution 200 g NaOCl = 10^{6}g solutiong NaOCl 5.25 g NaOCl = 100 g Chloroxg Chlorox **Do the math.**Set up the conversion factors so that all units cancel to give g Chlorox. Use the roadmap we developed in the previous step.**Check the answer.**You know the Chlorox solution is much more concentrated than the dilute solution, so you expect the amount of Chlorox required to be small. The best way to decide if your answer is correct is to work the problem again using different logic (e. g. the quick solution above.) The next best way is to see if the answer is consistent with the given information.
### Notes- Percentage compositions for solutions are ambiguous. "5.25% NaOCl" may mean any of the following:
**percentage type****interpretation of "5.25% NaOCl"****notation**mass/mass 5.25 g of NaOCl are dissolved in every 100 g of solution5.25 (w/w)% NaOCl mass/volume 5.25 g of NaOCl are dissolved in every 100 mL of solution5.25 g NaOCl/100 mL solution volume/volume 5.25 mL of NaOCl are dissolved in every 100 mL of solution5.25 (v/v)% NaOCl **not**equivalent. It's best to avoid mass/volume percentages entirely.The calculations assume that 5.25% NaOCl is a mass percentage rather than a volume percentage. A volume percentage is less likely because hypochlorite bleaches are produced by the reaction Cl and estimating the volume of NaOCl(aq) produced would be much less convenient than estimating its mass._{2}(aq) + 2 NaOH(aq) = NaCl(aq) + NaOCl(aq) + H_{2}O()
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