Chemistry
posted by Brian on .
If I add 0.348 liters of 5.25% bleach(Sodium Hypochlorite) into 4 liters of water, what is my concentration in ppm ?

I guess this is w/w

I think I got it  Please verify
Given:
c1v1=c2v2
MW of Sodium Hypochlorite = 74 grams/mol
5.25% = 52,500ppm = 52500 mg/l
calculating c1:
((52500mg/l)*(1mMol/74mg))/ (1liter/1000ml) = 0.709 mol/l or 0.709 M
rearranging c1v1=c2v2 for C2 gives:
C2= (0.709*0.348)/4 = 0.062 mol/l or 0.062M
Coverting 0.062 M to ppm
(0.062 mol/l)(1000mMol/mol)(74mg/mMol) = 4588 mg/l = 4588 ppm
Converting to %
4588ppm *(1%/10,000ppm)= 0.4588% Sodium Hypochlorite Solution
How'd I do? 
I followed to this step and stopped.
The final volume is 4.348 (assuming the volumes are additive) and not 4. The problem states that 0.348 L is ADDED to 4L.
rearranging c1v1=c2v2 for C2 gives:
C2= (0.709*0.348)/4 = 0.062 mol/l or 0.062M
Additionally I think there are some rounding errors here and there.
I would stick with the solution in my earlier post.