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I have a test result that I am trying to fully grasp. I need to determine how much of a contaminant I need to reach that amount in 4,800 gallons. Making the conversion to liters is no problem, nor is the math necessary to figure how much I need by weight to get this much contaminant.

The one thing I am unclear on is my assumption that $44.5~\mathrm{mg~L^{-1}}$ is by weight. In other words, if I have a result that measures $50~\mathrm{mg~L^{-1}}$ of a metal (nickel, for instance), that means if I were to be able to extract the metal and weigh it, it would weigh $50~\mathrm{mg}$.

Any help you can offer would be greatly appreciated.

bon
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user17960
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3 Answers3

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$50~\mathrm{mg~L^{-1}}$ is an expression of concentration by mass. It means that if you took one litre of the solution and extracted all the metal, the resulting metal would weigh 50 milligrams.

The other common option is to express concentration in terms of moles per litre ($\mathrm{mol~L^{-1}}$) which means that if you extracted all the solute from one litre of solution, you would have 50 moles of solute.

bon
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    Weight technically refers to a force exerted by a given mass, while the use of this term for masses is quite common, scientifically it is not correct. So the resulting metal has a mass of 50mg. – Martin - マーチン Aug 07 '15 at 06:43
  • Very true. Edited the answer to reflect that. – bon Aug 07 '15 at 08:42
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The test result is mass per volume, or concentration. It just says there are 44.5 mg of (whatever) in each liter of solution.

If you're asking how much total contaminant there is in 4800 US gallons given a test result of 44.5 mg/L: $$\left(44.5\,\mathrm{mg\over L}\right)\cdot\left({1\,\mathrm{L}\over 0.264172052 \,\mathrm{US\,gal}}\right)\cdot(4800\,\mathrm{US\,gal})\cdot\left({1\,\mathrm{g}\over 1000\,\mathrm{mg}}\right) = 808.6\,\mathrm{g}$$

Todd Minehardt
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In the given context, the quantity value $44.5\ \mathrm{mg\ l^{-1}}$ corresponds to the quantity mass concentration.

The mass concentration of substance $\ce{B}$ (recommended symbol: $\rho_{\ce{B}}$, alternative symbol: $\gamma_{\ce{B}}$, mass concentration of water: $w$) is defined as $$\rho_{\ce{B}} = m_{\ce{B}}/V$$ where $m_{\ce{B}}$ is the mass of substance $\ce{B}$ and $V$ is the volume of the mixture. The dimension of the mass concentration is $$\dim \rho_{\ce{B}} = \mathsf{L}^{-3}\;\mathsf{M}$$ The coherent SI unit for mass concentration is ‘kilogram per cubic metre’ (unit symbol: $\mathrm{kg/m^3}$).

In a different context, however, the quantity value $44.5\ \mathrm{mg\ l^{-1}}$ might as well correspond to the quantity mass density (or density).

The density $\rho$ is defined as $$\rho = \mathrm dm/\mathrm dV$$ where $m$ is mass and $V$ is volume.

For example, the density of hydrogen at a temperature of $T = 20.0\ \mathrm{^\circ C}$ and a pressure of $p = 0.538\ \mathrm{bar}$ is $\rho = 44.5\ \mathrm{mg/l}$.

In general, quantities of the same dimension are not necessarily of the same kind. Therefore, the unit symbol should not be used to provide specific information about the quantity, and should never be the sole source of information on the quantity. In plain language: it is important not to use the unit alone to specify the quantity.

  • I don't quite understand this formula: $\dim \rho_{\ce{B}} = \mathsf{L}^{-3};\mathsf{M}$ Is it $\mathsf{L}$ for a length and $\mathsf{M}$ for a mass? – Martin - マーチン Sep 20 '15 at 14:16
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    @Martin-マーチン yes, the dimensions (represented by a single sans-serif capital) of the base quantities in the SI and ISQ are $\mathsf{L}$ (length), $\mathsf{M}$ (mass), $\mathsf{T}$ (time), $\mathsf{I}$ (electric current), $\mathsf{\Theta}$ (thermodynamic temperature), $\mathsf{N}$ (amount of substance), and $\mathsf{J}$ (luminous intensity). Thus the dimension of any quantity $Q$ can be written in the form of a dimensional product $\dim Q = \mathsf{L}^\alpha;\mathsf{M}^\beta;\mathsf{T}^\gamma;\mathsf{I}^\delta; \mathsf{ \Theta}^\epsilon;\mathsf{N}^\zeta;\mathsf{J}^\eta$ –  Sep 20 '15 at 14:33