A stock solution containing Mn2 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:

1. For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL.

2. For solution B, 10.00 mL of solution A was diluted to 250.0 mL.

3. 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.

Answer:- 1.534g  Mn + HNO3 →→ Khal Volume = 1,0006 1504g (lam Mn/3 1,504 Mn 54.949 Mr.) =0.02883 mel mu

M₁ = md = 0.02863md Muta L 1,000 2 = 0,02883M Mn+ 2 = M,

M₂ = M.V₁/v2 = (0.02883 m) (32,00ml) = – (1000,0 ml) 0.001.4421 minta (A 1,442×10-3

M₂ = 10,0014412M) (10.09) (250.0 ml)=  0.0000576BM Mnta (B 5.766×10-5 My

= (0100005760M) (10.00me) (500.0me) = 0.000001154 MY Muta (@ 1.154×10-4