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March 31, 2015

March 31, 2015

Posted by **Chris** on Wednesday, September 27, 2006 at 10:20pm.

our molarity of the solution was .100 +/- .6%. We made a stock solution of .5836 g of NaCl/ 100 mL of water. We then measured out 10. +/- .02 mL of solution with a transfer pipet and evaporated the water.

Our calculation looks like this:

10. +/- .02 x (.100 (sub 0) mol +/- .6%/ 1L) x (1 L/1000 mL) x (58.44 g/mol) = .58 +/- .005 g.

Is this correct for calculating the theoretical amount of NaCl?

58.36 g/L of soln = 1M and that is that same as 0.5836 g/100 mL solution (you say 100 mL water but that isn't the same as 100 mL of solution).

If you take 1/10 of that (you took 10 mL and that is 1/10 of 100), then the aliquot you transferred to the evaporating dish must be 1/10 x 0.5836 = 0.05836. If your number of 0.58 g is correct, then you expect to find in the 10 mL what you started with in the 100 mL. Doesn't sound right to me. BUT I just ran through you arithmetic and obtained 0.058. Check your arithmetic. Check my thinking.

That does make sense, I was wondering why my calculations were yielding the same amount of NaCl with the 10. mL as the 100. mL. I might have made a calculation error. Thank you.

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