Dilution of Solutions Problem
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#1
September 20th, 2006, 16:00
 ConfuzeredN Junior Member Join Date: Sep 2006 Posts: 1
Dilution of Solutions Problem

After being stuck on this ridiculous problem for about an hour and a half... (im on a 3 wk break and my grade is slipping so im taking my time) I have come for help. This problem comes from my Raymond Chang Chemistry 7e textbook. Chapter 4 number 70. Please help!

4.70) A 46.2ml, 0.568 M calcium nitrate solution is mixed with 80.5 ml of 1.396M calcium nitrate solution. Calculate the concentration of the final solution.

I know the answer is 1.09 M and I've tried just about anything, but i can't seem to get there!
#2
September 21st, 2006, 21:37
 yhtu22 Junior Member Join Date: Sep 2006 Location: San Marcos, TX Posts: 2
got it

its not too hard. You are adding two volumes of the same solution together. SO at the end you have to divide by the new total volume.

First calculate the mols for each solution. Then add them up. Next divide that value by the total volume. Like this:

.0462L (.568mol/L) = .026242mol

.085L (1.396 mol/L) = .11866 mol

.026242 mol + .11866 mol = .144902 mol (total mols in solution)

.0462L + .085L = .1312L (total volume of solution)

Mol/L = molarity(M) = .1144902mol / .1312L = 1.1044

Just remember that you have to deal with the total volume of the new solution. As well as the total amount of mols...
Hope that helped.

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