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A batch of 1,000 kg of KCl is dissolved in sufficient water to make a saturated solution at 363 K, where the

solubility is 35 wt % KCl in water. The solution is cooled to 293 K, at which temperature its solubility is 25.4 wt %. a) What are the weight of water required for the solution and the weight of KCl crystals obtained? b) What is the weight of crystals obtained if 5% of the original water evaporates on cooling?

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Answer: hare first app to calculate the amount of water present in the solution to make 35% KCL solution. Once you get the... View the full answer

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Answer :
at 363K, Solubility is 35 Wty.
35 why. Kel =
1000 k8 of kel
X 10 0
x ks of solution
XK8 SoIn =...
1000 K8 x100
35
= 2837.14 k8
". weight of water = solution weight (x k8) - weiset of...

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