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epherbertson:
Hi
I have been struggling with this one for a while (probably super easy). Any help is appreciated



Thanks

kiwiberry:

--- Quote from: epherbertson on March 23, 2017, 03:28:04 pm ---Hi
I have been struggling with this one for a while (probably super easy). Any help is appreciated

(Image removed from quote.)

Thanks

--- End quote ---

We know that m(PbCl2) = 0.595 g
Therefore n(PbCl2) = 0.595 / (207.2+2*35.45) = 0.21395... mol
n(PbCl2):n(Pb2+) = 1:1
So n(Pb2+) = 0.21395... mol
m(Pb2+) = 0.21395...*207.2 = 0.44330... g
c(Pb2+) = 0.44330... / 0.05 = 8.866... g/L
So the answer is A! :)

epherbertson:
Thank you so much
I literally did that exact same working, must have just put the numbers in my calculator wrong.
Thanks heaps

mylinh-nguyen:

--- Quote from: kiwiberry on March 23, 2017, 04:34:48 pm ---
We know that m(PbCl2) = 0.595 g
Therefore n(PbCl2) = 0.595 / (207.2+2*35.45) = 0.21395... mol
n(PbCl2):n(Pb2+) = 1:1
So n(Pb2+) = 0.21395... mol
m(Pb2+) = 0.21395...*207.2 = 0.44330... g
c(Pb2+) = 0.44330... / 0.05 = 8.866... g/L
So the answer is A! :)

--- End quote ---

quick question if this is ever seen, why do you divide mass over volume when finding conc of lead
m(Pb2+) = 0.21395...*207.2 = 0.44330... g
c(Pb2+) = 0.44330... / 0.05 = 8.866... g/L

kiwiberry:

--- Quote from: mylinh-nguyen on July 16, 2017, 05:25:33 pm ---quick question if this is ever seen, why do you divide mass over volume when finding conc of lead
m(Pb2+) = 0.21395...*207.2 = 0.44330... g
c(Pb2+) = 0.44330... / 0.05 = 8.866... g/L

--- End quote ---

Here you need to find the concentration in g/L, so you divide mass by volume. If you wanted to find it in mol/L, then you would use the formula c=n/v :)

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