NaClO Concentration

In my previous occupation the usage of bleach--$NaClO$--I found was quite staggering, and based off of the health issues that I and other coworkers were having, I wanted to see how close we were cutting it to being over exposure limits. Taking the definition, $$1 \, \, ppm = \frac{1 \,\,mg \,\,solute}{1 \,\,L \,\,solvent} = \frac{1 \,\,g \,\,solute}{1 \,\,m^3 \,\,solvent}$$

I have a $946 \,\,mL$ bottle of solution labeled to be $.65\%$ Sodium Hypochlorite. This leads me to calculate the presence of $6.149 \,\,mL$ of the entire mixture being $NaClO$. Subtracting that from the total volume to get the volume of the solvent, I obtain $.940 \,\,L \,\,solvent$. Using the fact that the density of $NaClO$ is $\rho =1.11 \frac{g}{mL}$, I obtained $6.83\,\,g$ or $6830\,\,mg\,\,solute$. Plugging these values in for the concentration in $ppm$ I obtain $\frac{6830\,\,mg}{.940\,\,L}= 7265.957\,\,ppm$ ($\frac{g}{m^{3}}$).

The American Industrial Hygiene Association only recommends $2\,\,ppm$ for short term exposure, which leads me to believe that my calculations might be in error in some way. Is this calculation correct or do I have some fundamental misunderstanding about the concept of molarity? I just don't feel like something that's $3633$ times more concentrated than the AIHA recommends wouldn't even make it off the drawing board, let alone into workplaces.

• The "0.65%" you refer to is ambiguous. You took to mean volume percent, but it could mean mass percent. The would mean the solution was 6500 ppm sodium hypochlorite. It could also mean the notorious and misleading "% weight-volume", i.e., it could mean 6.5 g/L. In any case the exposure limits are for airborne concentrations, not for liquid phase concentrations, as Loong's answer correctly notes. – Curt F. Mar 26 '15 at 20:30