Chemistry

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

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  1. I guess this is w/w

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  2. 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?

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  3. 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.

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