CHEMISTRY HELP!! Dilution

A 3.00 gram of solid Na2CO3 molar mass 106 g/mol, was dissolved in sufficient water to make 250 mL of solution. A 10.0 mL sample of this solution was diluted 100 mL What is the concentration of Na+ in this new solution?

A. 0.113 M
B. .143 M
C. .0226 M


Please this was on my last exam I know the answer was C... but can someone explain to me how to do this because I really don't understand the method... I always thought dilution problems were just M1V1=M2V2
My chem final is in two days

thanks so much <3

You want the concentration of Na+, not the concentration of Na2CO3

Each Na2CO3 dissociates to give 2Na+ ions
Na2CO3 -----> 2Na+ + CO3^2-

So the concentration of Na+ is 2 x the concentration of Na2CO3

moles Na2CO3 = mass / molar mass = 3.00 g / = 0.0283 mol
moles Na+ = 2 x moles Na2Co3 = 0.05561 mol

Molarity Na+ = 0.0556 mol / 0.250 L = 0.266 M

Now you can use V1M1 = V2M2

M2 = 10 ml x 0.266 M / 100 ml
= 0.0266 M

Moles = 3 / 106 = 0.0283 moles
You took out 10 /250 x 0.0283 moles =0 00113 moles
M= 0.00113 / 0.1L =0.0113 M
Use common logic for this one.