I wouldn't go that route. I think this route is easier to see.
First, determine the molarity of the inital formaldehyde solution.
Thats 1.037 g/mL x 1000 mL x (37/100) x (1 mol/30 g) = about 12.8 M
Then use the dilution formula of
mL1 x M1 = mL2 x M2
For the first dilution you have
1 mL x 12.8 = 500 mL x M2
and M2 = 1*12.8/500 = approx 0.03 but you need a more accurate answer than that.
For the second dilution you have to assume that the volumes of additive; i.e., that 1 mL + 4 mL = 5 mL, then
1 mL x 0.03M = 5*M2 and solve for M2.
Another way to do the dilutions is
12.8 M x (1/500 ) = about 0.03
and 0.03 x 1/5 = ?
A third way is to look at the dilution factor.
You start with 12.8 M
You dilute 1 to 500 and again 1 to 5 so the total dilution is 500*5 = 2500 so the final concn is 1/2500 th the initial or 12.8/2500 = ?
A liquid solution of formaldehyde is 37% pure. This solution has a specific gravity of 1.037. The molecular weight of formaldehyde is 30.0. To 1.00 ml of this impure stock solution is added quantity sufficient total the total volume to 500 ml. If 1.00 ml of this solution is added to 4 ml of water in a final reaction, what is the molarity of formaldehyde in this final reaction tube?
So far: I have found the grams by doing
.001 L * 1.037 * .37 g/L = 3.84*10^-4
Then I found the M from here
3.84*10^-4 g * 30 moles/g = .0115 moles/ 1L = .0115 M
I don't thing this is correct because I didn't use any dilution or anything? Some direction on where I would use the dilutions would be helpful
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