Chem
posted by Lena on .
As a part of a study to see whther just the chloride ion on NaCl accelerates the rusting of iron, a student decided to see whtjer iron rusted in a solution on Na2SO4. The solution was made by dissolving 4.550 g of Na2SO4 in water using 125.0 mL volumetric flask to adjust the final volume. What is the molarity of the solution?
I don't understand this question at all. What do you do with the iron :S

Actually, the question isn't stated very well but what they mean is that 4.550 g Na2SO4 are dissolved in a total volume of 125 mL. They want you to calculate the molarity. To do this, just remember the definition of molarity. It is the number of moles per liter of solution.
So you use 4.550 g Na2SO4, calculate the moles in that mass, then moles/L (the L is 0.125 ) gives the molarity. 
mass = 4.550 g
molar mass = 142.05 g/mol
n= 0.032030975 mol
c= 0.032030975/0.125 = 0.256 mol/L
would I have 3 or 4 sig. digs? cause when I converted 125.0mL into 0.125 L I lost a sig dig;or should it be 0.1250 L? 
You have 4 s.f. in 4.550.
There are 6 s.f. in 142.043 which I used instead of 142.05).
There are 4 s.f. in 125.0 mL (or 0.1250).
Therefore, you are allowed 4 s.f. (from the lowest which is 0.1250 or 4.550) so the answer is 0.25626 moles/L which rounds to 0.2563 M to 4 s.f. [I take issue, somewhat, that you lose one place when converting 125.0 to 0.1250. I see 4 s.f. in both 125.0 as well as in 0.1250.] 
Thank you so much :)