A stock solution containing Mn2 ions was prepared by dissolving 1.584g of pure manganese metal in nitric acid and diluting to a final volume of 1.000L. The following solutions were then prepared by dilution:a. For solution A, 50.00mL of stock solution was diluted to 1000.0mL. b. For solution B, 10.00mL of solution A was diluted to 250.0mL. c. For solution C, 10.00mL of solution B was diluted to 500.0mL. Calculate the concentrations of the stock solution and solutions A, B and C.3

The concentration of the stock solution is 0.029 M

The concentration of the solution A is 0.00145 M.

0.000058 moles/litre is the concentration of Solution B.

0.00000116 moles/ litre is the concentration of solution C.

Explanation:

Weight of the manganese metal dissolved is 1.584 and diluted in 1000 ml.

The number of moles of manganese will be calculated as

Number of moles = mass ÷ atomic mass  (atomic mass of manganese = 54.93 gram/mole)

Thus number of moles = 1.584 ÷ 54.93

                                        = 0.029 moles

Molarity or concentration of the Mn ions is calculated by the formula:

M= n ÷ V  

M = 0.029 ÷ 1

    0.029M is the molarity of the solution given.

Now to know the molarity of solution A

It can be known by

M1V1 = M2V2

0.029 × 50 ml = M2 × 1000 ml

   M2= 0.00145 M thus the molarity of the solution A is 0.00145 M.

Molarity of solution B  ( 10 ml of solution A is diluted to 250 ml)

Applying the formula:

M1V1 = M2V2

0.00145 × 10  = M2 × 250   ( ml will be converted to L by dividing the volume with 1000)

                             M2 = 0.000058 moles/litre is the concentration of Solution B.

For solution C ( 10 ml of solution B is diluted to 500ml)

From the formula:

M2 V2 = M3V3

0.000058 × 10 = M3 × 500 ( Volume will be changed to L )

0.000058 × 0.01 = M3 × 0.5

= 0.00000116 moles/ litre is the concentration of solution C.