Playing with Solubility 3

Chemistry Level 3

At 6 0 ∘ C 60^\circ\text{C} , a 150 g K N O X 3 \ce{KNO3} solution must add 22 g K N O X 3 \ce{KNO3} crystal or evaporate 20 g of water to become a saturated solution. If the 150 g K N O X 3 \ce{KNO3} solution contains 68 g of K N O X 3 \ce{KNO3} and I evaporate 30 g of water, what is the amount of K N O X 3 \ce{KNO3} crystal that be separated in grams? Assume that the temperature does not change during the process.


The answer is 10.8.

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Christopher Boo by Christopher Boo , 24, Singapore

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1 solution

If the 150 g K N O X 3 \ce{150 g \space KNO3} solution has 68 g \ce{68 g} of K N O X 3 \ce{KNO3} then there is 150 − 68 = 82 g 150 - 68 = \ce{82 g} of water.

Then the solubility of K N O X 3 \ce{KNO3} at 6 0 ∘ C 60^\circ C is 68 82 − 20 = 68 62 = 34 g 31 g \dfrac{68}{82-20} = \dfrac{68}{62}= \dfrac{34 \text{ g}}{31 \text{ g}} of water.

The mass of K N O X 3 \ce{KNO3} , 82 − 30 = 52 g 82-30 = 52 \text{ g} of water can dissolved at 6 0 ∘ C 60^\circ C is 34 31 × 52 = 57.03 g \dfrac{34}{31} \times 52 = 57.03 \text{ g} .

Therefore, the amount of K N O X 3 \ce{KNO3} crystals are separated out is 68 − 57.03 = 11.0 g 68 - 57.03 = \boxed{11.0} \text{ g} .

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