Q1: A stock solution containing Mn+2 ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution: For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, 10.00 mL of solution B was diluted to 500.0 mL. Calculate the concentrations of the stock solution and solutions A, B, and C.

The concentration of a solution is the amount of solute in a solution.

We have from the question that he mass of manganese = 1.584 g

Hence;

Amount of = 1.584 g/55g/mol = 0.0288 moles

Recall that;

Number of moles = concentration * volume

Let the concentration of the solution be C

0.0288 moles = C * 1 L

C = 0.0288 moles/ 1 L

C= 0.0288 mol/L

Hence concentration of stock solution = 0.0288 mol/L

For solution A

From the dilution formula;

C1V1 = C2V2

where;

C1 = initial concentration

C2 = final concentration

V1 =  initial volume

V2= final volume

C1 = 0.0288  mol/L

V1 = 50.00 mL

C2 = ?

V2 = 1000.0 mL

C2 = 0.0288  mol/L * 50.00 mL/1000.0 mL

C2 = 0.00144 mol/L

Hence, concentration of solution A is 0.00144 mol/L

For solution B

C1V1 = C2V2

C1 = 0.00144 mol/L

V1 = 10.00 mL

C2 = ?

V2 = 250.0 mL

C2 = 0.00144 mol/L * 10.00 mL/250.0 mL

C2 = 0.0000576 mol/L

Hence, concentration of solution B is 0.0000576 mol/L

For solution C

C1 = 0.0000576 mol/L

V1 = 10.00 mL

C2 = ?

V2 = 500.0 mL

C2 = 0.0000576 mol/L * 10.00 mL/500.0 mL

C2= 0.000001152 mol/L

Hence, concentration of solution C is 0.000001152 mol/L

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