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Topic: What is the resulting molarity?  (Read 5860 times)

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What is the resulting molarity?
« on: March 28, 2006, 07:00:09 PM »
I'm going crazy over this question that I can't complete :'(
The question reads: You prepare a "stock" solution by dissolving 50.0 grams of potassium chloride (KCl) in enough water to make 250mL of solution. If you place 100mL this stock into a beaker and then add (another) 50mL of water to it, what is the resulting molarity?

I figure that I need to find the molarity of the "stock" solution:
My work: Molar Mass of KCl = 74.55g/mol
50g/74.55 = .67moles
250mLx1l/1000mL = .25L ----> 1000/250 = 4
4x.67/.25L = 2.68mol/L

I can't figure out where to go after that. Do I plug the 2.68mol/L to a different equation?

Offline Yggdrasil

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Re:What is the resulting molarity?
« Reply #1 on: March 28, 2006, 10:22:12 PM »
For dilutions, you can use the following equation:

M1V1 = M2V2

Where the Ms refer to the molality of the solutions and V refer to the volumes before and after dillution.

So in this case, your M1 is 2.68M and your V1 is 100mL.  When you add 50mL of water, the final volume of your solution is (approximately )150mL so V2 is 150mL.  Simple algebra gets you to the answer of M2 = 1.79M.

Offline Borek

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Re:What is the resulting molarity?
« Reply #2 on: March 29, 2006, 03:32:45 AM »
Alternatively think this way: your stock solution contains 50g KCl in 250 mL, you are using 100mL - so you took 100/250*50 = 20g KCl and dissolved it in 150mL (which is the final volume - 100+50).

Both ways are correct, Yggdrasil's is more formal, my probably faster :)
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Re:What is the resulting molarity?
« Reply #3 on: March 30, 2006, 07:12:10 AM »
I just wanted to say a big ole "Thank You" for your help with this equation. It was much appreciated by not only me but by my classmates that shared the same dilemma. ;D

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