Moles/concentration problem?

Question

25.6g of anhydrous sodium carbonate is dissolved in 200mL of water.

(i've found that concentration in mol/L = ~1.21)

A 20mL sample of this solution was placed in a 150mL flask and 80mL of water added. Determine the concentration of sodium ions in his diluted solution.

(That is the *exact* wording of the question from the book - its a bit hard to understand, imo)

Any help with this is appreciated :]

The answer: 0.484 mol L-1

Answer 1

20 mL x 1 L/1000 mL x 1.21 mol/L=0.0242 mol Na2CO3

20 mL+80 mL=100 mL=0.100 L

0.0242 mol/0.100 L=0.242 M Na2CO3

0.242 M Na2CO3=2*0.242=0.484 M Na+

Because there's 2 Na+ ions for every 1 M of Na2CO3 you have to multiply by 2 to get the answer.  

Answer 2

M =  25,6 X 5 / 106

    =  1,208 mol / l

So 1,21 is correct

Diluted solution strength is 1,208 X( 20 / ( 80 +20))

= 0,242 mol / l sodium carbonate

= 0,484 mol / l  Na+

Answer 3

Assuming you calculated the concentration correctly, then the following would hold:

20 ml x 1.21 M = 100 ml (X M)

X = 0.242 M Na2CO3

Since there are 2 moles of Na per mole of Na2CO3, the answer comes out 2 x 0.242 = 0.484 mol L-1

Answer 4

Hippy