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Symptoms of mercury poisoning become apparent after a person has accumulated $\pu{20 mg}$ of mercury. If a person ingested $\pu{30 mg}$ of mercury, what concentration of mercury in parts per million, are in his body? Assume the person has a mass of $\pu{63.5 kg}$."

My thought process, using dimensional analysis, was to convert the amount of $\pu{mg}$ the person ingested into $\pu{g}$. Then I took it ($\pu{0.03 g}\ \ce{Hg}$) and divided by the molar mass of mercury to find the number of moles of mercury there are.

I then divided the amount of mercury by $\pu{63500 mL}$ because I assumed $\pu{1 g} = \pu{1 mL}$. I did not use the $\pu{20 mg}$ information that it got me at all. Thus what I got was:

$$\pu{0.03 g} \cdot \frac{\pu{1 mol}}{\pu{200.59 g}} \cdot \frac{1}{\pu{63500 mL}} = \pu{2.3e-9 mol L-1}$$

However, I completely messed up, because first, my answer was in $\pu{mol L-1}$, and I'm not sure if you're supposed to use the molar mass of mercury here. The answer was $\pu{0.47 ppm}$, but I don't know how to get to here. Could someone explain the thought process for this?

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There are 1,000,000 mg in a kg (this is, mg/kg = ppm), so all you need to do is divide the mass of $\ce{Hg}$ by the mass of the person as-is (the units work out to ppm):

$$\left({30\,{\rm mg}\over 63.5\,{\rm kg}}\right)\cdot\left({{\rm ppm}\over {\rm mg\cdot kg^{-1}}}\right) = 0.47\,{\rm ppm}\,\ce{Hg}$$

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  • $\begingroup$ Where did mg x kg^-1 come from? $\endgroup$ – CrazyNinja A Nov 18 '17 at 21:06
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    $\begingroup$ mg/kg is exactly mg x kg^-1, of "millgram divided by kilogram". Please upvote and accept the answer if you find it helpful. $\endgroup$ – Todd Minehardt Nov 18 '17 at 22:03

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