Question 376377: Hi!
I'm working on a chemisty problem, for which I have the answer, but can't seem to work the equation correctly and am thinking the answer the instructor gave us is incorrect. Please help!! What volume of water is needed to prepare 500.0 mL of a 0.250 M Ca(NO3)2 solution from a 5.00 M Ca(NO3)2 solution? I'm using (M1)(V1)= (M2)(V2)..I changed my mL to L = 0.5 L, then (0.250)(5.00)=(0.5)(V2), then 1.25/0.5 = V2, then V2 = 2.5...The answer from my instructor is 475 mL...Am I wrong? Thanks!!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let
M1 = molarity of the final solution, V1 = volume of the final solution (in mL), M2 = molarity of original solution, and V2 = volume of original solution (in mL).
So this means that
M1 = 5, V1 = x, M2 = 0.25, and V2 = 500
Plug all of this into  to get the equation  which becomes  . Solve for x to get 
So because V1 is a volume in mL, the volume of the final solution is 25 mL
So we have 25 mL of  , which means that there must be  mL of water left over.
Note: I'm not a chemistry specialist so my terminology might be a bit off, but the answer, and the steps taken to get the answer, are correct.
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