ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: zibb3r on February 14, 2011, 10:37:05 pm
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Hi Guys,
Q: Use the molar masses of barium and sulfate to calculate the mass of sulfate ions in the Barium Sulfate solution?
A (by vea {thanks vea!}): BaSO4---->Ba2+ + SO4 2-
So the mole ratio for BaSO4 and SO4 2- is 1:1.
n(SO4 2-)=n(BaSO4)=0.0327mol
m(SO4 2-)= no. of mole X molar mass
=0.0327 X (32.07+64.00)
=3.141g (4 significant figures)
Would the question be asking the same thing if the question was: Use the molar masses of barium and sulfate to calculate the mass of sulfate ions in the precipitate formed? Or Not??
Thanks Everybody for your help!!!! :) :smitten:
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No worries zibb3r :)
I guess I'll leave someone else to check my working...
As for your question:
1. Use the molar masses of barium and sulfate to calculate the mass of sulfate ions in the Barium Sulfate solution?
This question asks for the mass of sulfate ions in the barium sulfate solution- the keyword here is solution.
2. Use the molar masses of barium and sulfate to calculate the mass of sulfate ions in the precipitate formed?
This question asks for the mass of sulfate ions in the precipitate formed. Since the precipitate would be part of the solution (consisting of precipitate and other ions) it is essentially the same question.
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yep thats right :)