Hello, I am currently in the first year of my chemical technician training and would need help with the following task:
Calculate the required portion of potassium nitrate solution to dissolve 1500g of KNO3 at 60°C. Then, allow for a 50% excess.
L(KNO3; at 20°C) = 31.7g/100g water
L(KNO3; at 60°C) = 109.9g/100g water
∆L(KNO3) = 78.2g/100g water
(1 votes)
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I don’t know what you’re talking about. But if 109.9 g of KNO3 dissolve in 100 g of water, you obviously need 1.36 kg of water to release 1.5 kg of KNO3.
Because I don’t really know what’s in demand, I just keep reckoning.
After cooling to 20° C., 100 g of water can only dissolve 31.7 g of KNO3 or 1.36 kg of water can only dissolve 433 g. So the difference crystallizes out, 1067 g.
If you are thinking of reducing 50% more water than necessary at 60° C., i.e. 2.05 kg, then the KNO3 naturally dissolves in it, and after cooling, the amount of water can still dissolve 649 g of KNO3, so you will get less material, i.e. 851 g — that is only 80% of what you would have been able to use if the 1.36 kg.