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Author: Subject: math/ chemistry question
jewels
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[*] posted on 28-10-2004 at 09:03
math/ chemistry question


I'm not sure how to set this question up..
How many milliliters of a 2.5 M MgCl2 solution contain 17.5g MgCl2

any help on this would be great!!
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David Marx
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[*] posted on 28-10-2004 at 09:26


Now I suspect that you are going to get flamed for posting yet another chemistry homework problem on the board but...

This is a really simple problem, like many you will be forced to deal with anywhere in chemistry/science/engineering. It may look tricky, due to the seemingly unrelated units. A good way to analyze these is with the "railroad tracks" method of dimensional analysis. Start by writing your starting solution on top of a line. Now draw a vertical line next to it. Next to this put down what the units are equivalent to on the top and the starting units on the bottom. Proceed to cross out the starting units, since they are in both the numerator and the denominator. Keep working conversions like this until you are left with the units you need for the final answer.

Don't get discouraged, just keep using all the conversions you are been taught. If you do it correctly, even if you do superfluous steps it will not matter and the answer will come out correctly.




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[*] posted on 28-10-2004 at 09:37
Thanks David


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[*] posted on 29-10-2004 at 08:39


The first step is to figure out what a Mole of MGCl2 weighs. Adding up the Atomic Weights of the formula you get: Mg=24.3 and Cl=35.4 but since there are two Chlorines you add it twice: 75.8 + 24.3 = 100.1 grams is one mole of Magnesium Chloride. This is 6.02 x 10^23 molecules of MgCl2, also known as Avagadro's number. You are told you have a solution that is a 2.5 Molar concentration, which means each Liter (1000 milliliters) contains 2.5 Moles or 100.1 x 2.5 =250.25 grams of MgCl2. The question then comes down to how many ml of this does it take to make 17.5 grams. I'm not going to do the final step but if you can set up a ratio of 1000ml:250.25gm= (x)ml:17.5gm you should be able to solve it for the volume.
I hope I didn't make any mistakes ;-)
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