The solubility of KNO_3 is 155 g per 100 g of water at 75 degree C and 38.0 g at

Question

The solubility of KNO_3 is 155 g per 100 g of water at 75 degree C and 38.0 g at 25 degree C. What mass (in grams) of KNO_3 will crystallize out of solution if exactly 100 g of its saturated a solution at 75 degree C is cooled to 25 degree C? A 50-g sample of impure KCIO_3 (solubility = 7.1 g/100 g H_2O at 20 degree C) is contaminated with 10 percent of KCI (solubility = 25.5 g per 100 g of H_2O at 20degree C). Calculate the minimum quantity of 20 degree C water needed to dissolve all the KCI from the sample. How much KCIO_3 will be left after this treatment? (Assume that the solubilities are unaffected by the presence of the other compound.)

Explanation / Answer

12.28) the solubility of KNO3 is 155 g / 100g of water at 750C

the solubility is 38 g / 100g of water at 250C

If the saturated solution is cooled from 75 0Cto 25 0C then

the solubility will decrease from 155 to 38 grams

So amount of crystallized salt = 155-38 = 117 grams

Example 16

A 50g os sample of impure KClO3 [solubility 7.1 g / 100 g of water]

contaminated with 10% KCl

10% KCl means the amount of KCl in 50 g of KClO3 = 0.1 X 50 grams = 5 grams

water needed for solubility of 5 grams of KCl will be calculated from its solubility ( solubility 25.5 g / 100g of water]

for 25.5g of Kcl = 100 g of water is needed

for 1 g of KCl = 100 /25.5 g of water will be needed

For 5 g of Kcl = 5 X 100 / 25.5 g of water will be needed = 19.61 grams of water

Now as the sample is 50 grams out of which 5 gram is KCl so KClO3 present = 45 grams

100 grams of water is needed for 155 g of KClO3

1 gram will dissolve = 155/100 g of KClO3

19.61 grams will dissolve = 19.61 X 155 / 100 grams of KClO3 = 30.395 grams of KClO3

so KClO3 left = 45 - 30.395 = 14.605 grams

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