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Topic: Saturated Solution Problem  (Read 12607 times)

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natalie_2006

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Saturated Solution Problem
« on: January 12, 2006, 11:50:37 AM »
 Hey I'm having trouble with this chemistry question... Can anyone help??

A chemist wishes to prepare 200mL of a saturated solution of BaF2, the concentration of which is 6.3 x 10-3 mol/L.
So how do I determine the mass of BaF2 that will be needed to make up the 200mL of the solution??

I just don't know where to begin....

Thanks!
« Last Edit: January 12, 2006, 11:54:07 AM by natalie_2006 »
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Offline Alberto_Kravina

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Re:Saturated Solution Problem
« Reply #1 on: January 12, 2006, 12:14:37 PM »
Quote
Hey I'm having trouble with this chemistry question... Can anyone help??

A chemist wishes to prepare 200mL of a saturated solution of BaF2, the concentration of which is 6.3 x 10-3 mol/L.
So how do I determine the mass of BaF2 that will be needed to make up the 200mL of the solution??

Concentration in mol/L = number of moles / Volume (c=n/V)
You can calculate how many moles (n) you have to dissolve to make a 200mL-satured BaF2 solution.
After that you can calculate the MASS of Barium fluoride with the formula n=m/M (M is the molar mass, you can calculate it out of the periodic table). Calculate the mass and that's it! :)
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natalie_2006

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Re:Saturated Solution Problem
« Reply #2 on: January 12, 2006, 04:58:04 PM »
Hey, thanks so much.
I believe i'm on the right track

So I would change the equation to n=c*v and getting the value of 0.0126 mol, therefore
12.6*10-4 mol

Then I take that value you and use the formula you said.
So the molar mass of Barium Fluoride is about 175.3

so change formula to  mass = number of moles multiplied by the molar mass

mass = (12.6*10-4 mol) (175.3)

so the mass needed is 2.20878 which rounded would be 2.21

Am I correct?
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kkrizka

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Re:Saturated Solution Problem
« Reply #3 on: January 14, 2006, 02:54:25 PM »
Quote from: natalie_2006 on January 12, 2006, 04:58:04 PM
Hey, thanks so much.
I believe i'm on the right track

So I would change the equation to n=c*v and getting the value of 0.0126 mol, therefore
12.6*10-4 mol

Then I take that value you and use the formula you said.
So the molar mass of Barium Fluoride is about 175.3

so change formula to  mass = number of moles multiplied by the molar mass

mass = (12.6*10-4 mol) (175.3)

so the mass needed is 2.20878 which rounded would be 2.21

Am I correct?

Seems like you converted from mL to L wrong. 200mL is .200L, so the equation is:
6.3x10^-3 * 0.200 = 0.00126

Otherwise is looks OK.
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natalie_2006

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Re:Saturated Solution Problem
« Reply #4 on: January 23, 2006, 08:05:39 PM »
You are right.. I was off cause of the exponents.
So it should be:
0.0063mol/L x 0.200L = 0.00126
therefore making it: 12.6 x 10-4

So is that correct then?  I'm just having trouble with the exponents when they are negative.
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