help me please?

Question

A stock solution containing Mn2+ ions was prepared by dissolving 1.588 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. Calculate the concentration of the stock solution.
____________ M

i got .0289.. it tells me im wrong....

The following solutions were then prepared by dilution.
(i) For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL. Calculate the concentration of solution A.
______________ M

(ii) For solution B, 10.00 mL of solution A was diluted to 250.0 mL. Calculate the concentration of solution B.
___________ M

(iii) For solution C, 10.00 mL of solution B was diluted to 500.0 mL. Calculate the concentration of solution C.
__________M

Answer 1

I get the same answer as you. My atomic weights table is 35
years old but I assume they haven't changed that much...
If the question is as you describe it then you are right. although perhaps 2.891E-2 Molar is "better"
Of course what "IT" is in your question is ? and what i"it" means by wrong is unknown.
Be sure to use the same number of significant digits (4) and go for it. 0.02891 moles per liter.
Also make sure you are making one liter. (I presume at STP)

Answer 2

Questions
1. divide weight/atomic weight of Mn, this will give you conc. Your number doesn't look all that bad.
i. For these, assume conc is 0.03 M
Problems follow formula.
V1M1 + V2M2 = (V1+V2)Mmix
v1=50, V2=950, M2=0, so 50x.03=1000Mmix
Mmix= 0.0015 M
2. Same deal, but here V1=10, V2=250, M1=0.0015M, M2=0.
3. Same deal again.