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.
To calculate the concentration of the stock solution, we first need to calculate the number of moles of manganese present in the solution.
To do this, we use the formula:
moles of Mn = mass of Mn / molar mass of Mn
The given mass of manganese is 1.542 g, but the atomic mass of manganese should have at least four significant figures. In this case, it is 54.94 g/mol.
Therefore, moles of Mn = 1.542 g / 54.94 g/mol = 0.028 mol
Next, we need to calculate the concentration of the stock solution, which is measured in moles per liter (M).
Concentration (M) = moles of solute / volume of solution in liters
The volume of the solution is given as 1.000 L.
Hence, concentration (M) = 0.028 mol / 1.000 L = 0.028 M
So, the concentration of the stock solution is 0.028 M.