Chemistry Purity Problems with solutions help

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The discussion centers on calculating the mass of calcium sulfate (CaSO4) that will dissolve in water, considering the purity of the sample. The solubility of CaSO4 at 25°C is 1.61x10^-2 mol/L, which translates to 1.0948 grams dissolving in 50.0 mL of water. The confusion arises from the interpretation of the purity percentage; the correct approach is to determine that 1.0948 grams of pure CaSO4 is required, regardless of the sample's purity. Thus, the user miscalculated the mass of the impure sample needed to achieve this solubility.

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loz123
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I have a test tomorrow and I thought I knew how to do these types of problems but my answers are all wrong.

Here is the question:
The solubility of CaSO4 at 25*C is 1.61x10^-2 mol/L. What mass will dissolve in 50.0 mL of water if the sample is only 20.0% pure?

Heres what I did. First I found moles of CaSO4:
n=CV
n=1.61x10^-2 x 0.05L
n=.00805 mol

Then I converted moles of CaSO4 to mass (grams):
m=0.00805mol x 136g/mol
m=1.0948g

Then I did a conversion of 100% pure to 20%:
1.0948g/100 = x/20
(Cross multiply)
100x = 21.896
x = .219g
Which means the sample of CaSO4 that is 20% pure should have a mass of .219g but according to the answers that is incorrect. Can someone tell me what I am doing wrong and how I can fix it. Thanks
 
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Welcome to PF, loz123! :smile:

I interpret your problem statement as that you need to find a total mass of which 20% is actually CaSO4.
As it is, 20% of .219 g is not 1.0948 g.
 
I find this question ambiguous - or intentionally tricky. You will dissolve 1.09g of CaSO4p/sub] in 50 mL of water regardless of the sample purity (unless sample contains other sulfates or other soluble calcium salt - but that's another problem).
 

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