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Topic: Molarity problem  (Read 4985 times)

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rsixtyone

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Molarity problem
« on: September 25, 2005, 12:32:09 PM »
hello all, i've tried this problem and i came up with 300 g

"A 40.0 mL portion of a 0.10M magnesium sulfate solution contains how many grams of magnesium sulfate?"

i go from volume to something and got lost a long the way. hmm...
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Offline Borek

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Re:Molarity problem
« Reply #1 on: September 25, 2005, 02:41:55 PM »
No, not 300 g.

Find number of moles, then convert to grams.
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rsixtyone

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Re:Molarity problem
« Reply #2 on: September 25, 2005, 07:55:01 PM »
i got .96 gram, is that correct?
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saibot

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Re:Molarity problem
« Reply #3 on: September 25, 2005, 08:13:46 PM »
I get 0.48 g as my answer.  Here is how I did it...

.10 mols/1000 ml = x mols/ 40.0 ml

x turns out to be 0.004 mols of MgSO4.  Converting mols to grams, we get 0.48.
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Offline mike

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Re:Molarity problem
« Reply #4 on: September 25, 2005, 08:30:17 PM »
Write down the data you know:

v = 40mL = 0.040L
c = 0.10mol.L-1
M = 121.34g.mol-1

the equations that link this data are:

n = c x v

m = n x M

First, the number of moles of MgSO4 in 40mL of 0.10M solution is:

n = 0.10 x 0.040 = 0.004moles

How many grams of MgSO4 is 0.004 moles?:

m = n x M = 0.004 x 121.34 = 0.485g
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