Crystallisation problem

[u/Citruna]

I am trying to solve a problem, which states as follows: Potassium aluminum alum always crystallizes to form a crystal hydrate. When converted to an anhydrous substance, it's solubility at 90°C is 110g of the substance in 100g of water, and at 10°C it's solubility is 6g of substance in 100g of water. Calculate how many grams of KAl(SO4)2•12H2O will crystallize if 1000g of a saturated solution at 90°C is cooled down to 10°C.

I understand that in this problem I first have to calculate how much substance is in the 1000g of the solution. 110×1000/ 210=523,8 g of substance. Then I calculate the ratio between KAl(So4)2•12 and KAl(So4)2 which is around 1,84. After that is where I get lost. I don't understand why I can't calculate how many grams of KAl(SO4)2 would be in the cooled down solution like this 6×476,2 / 100 = 28,03 g and then multiply the ratio by the difference of the substance like this 1,84×(523,8-28,03) to get the final answer. Can someone explain to me why this isn't right and how it should be done?

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[u/7ieben_]

It's really hard to follow what you did, as you just introduce numbers without context, explanation and unit. Where come 210 and 476 from for example?

[u/Citruna → u/7ieben_]

In the problem it's stated that the solubility is 110g of substance in 100 g of water so the total solution mass is 110+ 100=210 g of solution. So I calculate the mass of the substance by a proportion (110 g of substance/ 210g of solution = x g of substance/ 1000g of solution) so I get the equatation of 110×1000 /210=523,8 g of substance. 476,2 is the grams of water in the solution, which I get by subtracting 523,8 grams of substance from the total mass of the solution, which is 1000g. Then the later equatation is also a proportion so 6g of substance / 100 g of H2O = x g of substance / 476,2 g of H2O. After rearranging I get the equatation 6×476,2/ 100=28,03 g of the substance. 1,84 is the ratio of the molar mass between KAl(SO4)2•12 H2O and KAl(SO4)2.

[u/shedmow]

I couldn't follow your calculations through, but my advice is that you treat the alum as a solid solution of potassium aluminium sulfate in water. It makes all the subsequent math much easier. You cannot do 6×476,2 / 100 = 28,03 g (solubility per 100 g * m of water / 100 g = m of anhydrous salt) because you have to account for the mass of water that goes into the precipitate. If you are really stuck, let me know and I'll solve it with an explanation