Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Lenoon on January 28, 2020, 03:00:59 PM
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“To recrystallise a chemical a scientist adds it to 250g of water. 42.2% of it’s mass makes up the saturated solution at 80degrees. It is cooled to 20degrees where there’s an 11.1% of the chemical mass in solution. What is the mass of the precipitate of the chemical produced when this chemical is cooled from 80 to 20 degrees”
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It is a forum rule that you must show your attempt before we can help you.
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What do you think? There are two stages to the calculation.
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It is a forum rule that you must show your attempt before we can help you.
I just calculated the of the mass at 80 and 20 and then subtracted the results (105 and 27) and got 77 approximately and the answer is wrong I have no idea how to attempt it now
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What do you think? There are two stages to the calculation.
I have no clue man
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Its not very clear, but the first part of the question may be assuming that you filter the excess solid after making the saturated solution at 80C. Which means there would be no solid in the solution when cooling commences. So the 11.1% may be referring to the percentage of dissolved solute after filtration. Can you think of what it might be now?
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Is this the exact wording of the problem? There's a lot of ambiguity.
I read it as, if the scientist starts with X g of material, 42.2% of X is in solution (with 250 g water) at 80 C and 11.1% of X is in solution (250 g water) at 20 C. From this you can deduce the fraction of material that's in the precipitate at each temperature, and therefore the difference. But this would be in terms of X. To get an absolute number you'd need either X or the solubility of X at either temperature (since you're given the amount of water).
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Its not very clear, but the first part of the question may be assuming that you filter the excess solid after making the saturated solution at 80C. Which means there would be no solid in the solution when cooling commences. So the 11.1% may be referring to the percentage of dissolved solute after filtration. Can you think of what it might be now?
I’m not sure because would I just do 11.1% or 250g??
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You are told the final saturated solution contains 11.1% of the substance and mass of water is 250 g. It means the solution contains 250 g of water AND some amount of the substance. Lets say its mass is msubstance. So the total mass of the solution is 250+msubstance and (from the very definition of the mass percentage)
[tex]\frac{m_{substance}}{250 + m_{substance}}\times 100 \% = 11.1\%[/tex]
Can you now calculate mass of the substance left in the solution after precipitation? Before precipitation?