Chemistry II

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A stock solution containing manganese ions was prepared by dissolving 1.542 g pure manganese metal in nitric acid and diluting to a final volume of 1.000L. Calculate the concentration of the stock solution.

mols Mn = g/atomic mass Mn.
M = mols/L

1.542 g / 54.9 amu Mn = .028 mols Mn
0.28 mols Mn / 1.000 L = .028

1.542 g / 54.9 amu Mn = .028 mols Mn
0.28 mols Mn / 1.000 L = .028 M

Two errors.
1. The mass given is to four significant figures so you should use at least 4 for the atomic mass of Mn.
2.Check your arithmetic. 0.28/1.000 does not equal 0.028.


I see the problem with the math. You have 0.028 but you miscopied it as 0.28 on the next line. You should have four places in the answer, also.

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