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I am trying to do a dilution where I know I have 0.749 g/ mL and I want to use this concentration to calculate the grams of aspirin in each 100 mL solution. I am specifically worried about the units. I want grams out. Is it valid to do the following? One: $$C1V1=C2V2$$ $$(0.749 g)(1 mL)=(C2 g)(100mL)$$ $$C2=0.00749 g$$

or must I do this? Two: $$C1V1=C2V2$$ $$(0.749 g/mL)(1 mL)=(C2 g/mL)(100mL)$$ $$C2=0.00749 g/mL$$

The difference being the final units. g vs. g/mL

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    $\begingroup$ Concentration is g/mL so the latter is correct. $\endgroup$
    – DHMO
    Jan 28, 2017 at 5:24
  • $\begingroup$ So there is no way to get rid of the mL part? Part of me says to multiply by 1 mL to remove it... It seems like that wouldn't be valid though.... $\endgroup$
    – Math4Life
    Jan 28, 2017 at 5:33
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    $\begingroup$ What do you mean "to get rid of"? That's the way it is; you can't just up and change that, much like you can't get rid of the letter "c" in the word "concentration". Unless, of course, you want to know how much of your solute is in 1 mL (or 2 mL, or any other volume) of your solution; then you multiply by that volume and get some grams. $\endgroup$
    – Ivan Neretin
    Jan 28, 2017 at 6:28
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    $\begingroup$ You have to realize that if you have $$x_1 \cdot y_1 = x_2 \cdot y_2$$ then whatever the units of $x$ and $y$ the units will cancel appropriately. That doesn't mean that the closing price of the NY stock exchange times the number of Yankee's tickets sold means anything. $\endgroup$
    – MaxW
    Jan 28, 2017 at 6:40

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Concentration has units mass over volume; or particles (moles by convention) over volume.

It's fine to use $g/mL$ for concentration in a basic dilution equation.

$C$ is never a mass.

So your second equation has proper units.

If you want mass of aspirin in a given solution with concentration $C$, the mass is given by $m = C\times V$

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