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posted by  littlecaitling on 8/3/2008 7:06:45 PM  |  status: Live  

Dilution question

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Let's say I took 5mL of a 0.9730M stock Ni solution and added it to 20mL water.  What would the molarity of the resultant solution be?

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Sage
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posted by kwyjibo on 8/3/2008 7:21:35 PM  |  status: Live
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littlecaitling's comment:
"Thanks everyone, this is what I got too...I wanted to make sure I did it right"
Response Details:
5mL * (0.9730 M) = 4.865 M*mL (this is the amount of material, convert mL to L if you want to find the number of moles)

Then dilute so that total volume is 25 mL

4.865 M*mL/25 mL = 0.1946 mL

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posted by ClayL on 8/3/2008 7:23:24 PM  |  status: Live
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littlecaitling's comment:
"Thanks everyone, this is what I got too...I wanted to make sure I did it right"
Response Details:

Use the molarity to find the number of moles of Ni that you had:
.973M = x / .005L
.0049 mol Ni
Molarity is defined as the moles of solute per liter of solution, so the new molarity will be:
M = .0049 mol Ni / (.005+.02 L) = .196 Molar

Engineering ain't easy...
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posted by EmilyRua on 8/3/2008 7:37:04 PM  |  status: Live
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littlecaitling's comment:
"Thanks everyone, this is what I got too...I wanted to make sure I did it right"
Response Details:
Using the M1V1=M2V2 equation, and converting molarity into moles per liter, this turns into:
(0.9730 moles/1L)(5 mL) = (M2)(20 mL + 5 ml) then converting mL to L using 1 L= 1000 mL:
(0.9730 mol/L)(0.005 L)=(M2)(0.020 L + 0.005 L) then algebraic simplification and shuffling:
(0.9730 moles)(0.005 L) = (M2)
         (0.025 L)

the important thing to remember is that you are starting out with 5 mL and adding 20 additional, therefore you will have 25 mL diluted solution as your V2 quantity.
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