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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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tosey tosey
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11 years ago
A solution is prepared by dissolving 0.6901 g oxalic acid (H2C204) in enough water to make 100.0 mL of solution.  A 10.00 mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL.  What is the final molarity of the oxalic solution?
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wrote...
#1
11 years ago
c = n/v, where n = mass/molar mass.
then c1v1=c2v2
wrote...
#2
11 years ago
You start with a solution of 0.6901 gm oxalic acid.  Molar mass is 2 + 24 + 64  =  90

Then you take 10 ml of this solution, 10 ml will contain  0.06901 gm oxalic acid.

So you finish up with 0.06901 gm oxalic acid in 250 ml water. which would be 1.104 gm / lt.

Molarity  is 1.104/90  =  0.01227 molar.
wrote...
#3
11 years ago
molar mass of H2C2O4 = 90.0349 g/mol

moles H2C204 = (0.6901 g)(1 mole)/(90.0349 g) = 7.665 x 10^-3 mol

molarity of 1st solution = moles H2C2O4/Liters of solution = (7.665 x 10^-3 mol)/(0.1000 L)
molarity of 1st solution = 7.665 x 10^-2 mol/L

moles of aliquot = molarity x volume = (7.665 x 10^-2 mol/L)(0.01000 L) = 7.665 x 10^-4 mol H2C204

molarity of 2nd solution = (7.665 x 10^-4 mol)/(0.2500 L) = 3.066 x 10^-3 mol/L

The final molarity of the oxalic acid solution = 3.066 x 10^-3 mol/L
wrote...
#4
11 years ago
The 10mL aliquot contains 0.06901g oxalic acid
Dissolved in 250mL - That is 0.06901 * 4 = 0.27604 g per litre solution
 Molar mass oxalic acid = 90.04g/mol
0.2436g = 0.27604/90.04 = 3.066 *10^-3 moles dissolved in 1000 mL solution
 Molarity = 3.066*10^-3M
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