Question

this question has multiple parts. work all the parts to get the most points. a stock solution containing mn²+ ions was prepared by dissolving 1.060 g pure manganese metal in nitric acid and diluting to a final volume of 2.000 l. the following solutions were then prepared by dilution: for solution a, 40.00 ml of stock solution was diluted to 1500.0 ml. for solution b, 20.00 ml of solution a was diluted to 200.0 ml. for solution c, 20.00 ml of solution b was diluted to 550.0 ml. a calculate the concentration of the stock solution. molarity = m

Explanation:

Step1: Calculate the moles of Mn

The molar mass of Mn is \(M = 54.94\space g/mol\).
The moles of Mn, \(n=\frac{m}{M}\), where \(m = 1.060\space g\)
\(n=\frac{1.060\space g}{54.94\space g/mol}\)

Step2: Calculate the molarity of the stock solution

The volume of the stock solution \(V = 2.000\space L\)
Molarity \(M=\frac{n}{V}\)
Substitute \(n=\frac{1.060}{54.94}\) into the formula:
\(M=\frac{\frac{1.060}{54.94}\space mol}{2.000\space L}\)
\(M=\frac{1.060}{54.94\times2.000}\space mol/L\)
\(M=\frac{1.060}{109.88}\space mol/L\approx0.00965\space mol/L\)

Answer:

\(0.00965\)