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At 50 degrees on the solubility curve KClO3 should be saturated at 21g per 100g of water. Knowing this the one liter given to you in the question can be written as 1,000 ml or 1,000g of H2O since the density of water is 1g/ml. Set up an equation: 21g KCLO3/ 100g Water = x/ 1000g water

x=210g

Answer: 210g KClO3

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Q: How many grams of potassium chlorate must be added to 1 liter of water to produce a saturated solution at 50 degree celsius?
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