[coffee]

Determine the molarity of each of the following solutions:
1) 5.23g Iron (II) nitrate in 100.0cm3 (cube) of solution
Convert 5.23 g Fe(NO)2 to moles which would equal to 0.0451 mol Fe(NO)2
Then convert 100.0cm3(cube) to liters = 0.1 L
And finally 0.0451/0.1=0.451 mol/L

Did I do this right?

[Will]

Did I do this right?

Not quite- nitrate is NO₃^- so Iron(II) nitrate would be Fe(NO₃)₂.
Using accurate M_rs I worked out the number of moles in the solution to be 0.02907930365, and 0.291mol/L (3sf).

[coffee]

omg how did i forget the 3?!  >o<;; lol

Thanks for helping/replying. 

[coffee]

BTW, I got 0.315. how did you get a lower number then me? 

[Borek]

0.291M

You will not get more precise result...

CASC rulez

[Will]

BTW, I got 0.315. how did you get a lower number then me? 

How did you get 0.315? What did you work out the M_r of Fe(NO₃)₂ to be?

0.291M

Phew, it is nice to hear I got this one right from a reliable source.

[coffee]

o nvm i got what i did wrong. lol srry