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Topic: Concentration exercise  (Read 3166 times)

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Offline Evaldas

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Concentration exercise
« on: December 17, 2011, 11:11:04 AM »
Exercise: What volume of 3 mol/l NaCl solution (density 1.12g/cm3) is needed to add to 200 g of water in order to get a 10% solution?

If I understand correctly
200+x - 100%
x - 10%
to 200g of water we would need to add x=22.22g of NaCl to get 10% solution, but in this case we need to use a 3mol/l solution so I'm confused, how to calculate this....

Offline Arkcon

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Re: Concentration exercise
« Reply #1 on: December 17, 2011, 11:17:19 AM »
OK, can you use the definition of mol to convert mol/L into g per L?  Then you'll be able to calculate the volume needed.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Evaldas

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Re: Concentration exercise
« Reply #2 on: December 17, 2011, 11:21:13 AM »
Then 3 mol/l NaCl solution is 3x58.5g/1l=175,5g/l solution

Offline Evaldas

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Re: Concentration exercise
« Reply #3 on: December 18, 2011, 02:03:30 PM »
I still don't understand?

Offline Arkcon

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Re: Concentration exercise
« Reply #4 on: December 18, 2011, 03:13:02 PM »
So if there are 175,5g/l, what volume do you need to equal the grams you calculated before?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Evaldas

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Re: Concentration exercise
« Reply #5 on: December 18, 2011, 03:40:13 PM »
175.5 g - 1l
22.22 g - x l
x=22.22/175.5=0.126l?
But the thing is that now we have 200g of water plus 22.22g of NaCl and also y grams of water that is in the 0.126l 3 mol/l solution, so the solution is less than 20%. It would be 20% if we added 22.22g of pure NaCl to 200g of water.

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