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I have a liquid reagent with molecular mass $M = \pu{294.08 g mol^-1}.$ I'd like to prepare $V = \pu{400 ml}$ of solution of that reagent with a concentration $c = \pu{100 mM}.$

Does it mean I have to add

$$m = cVM = (\pu{0.1 mol l^-1})(\pu{0.4 l})(\pu{294.08 g mol^-1}) = \pu{11.76 g}$$

of reagent to $\pu{400.00 ml}$ of water, or should I add $\pu{11.76 g}$ of reagent to $\pu{388.24 ml}$ of water so that the total volume is $\pu{400 ml}?$

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2 Answers 2

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The amount concentration (molar concentration, molarity) is defined in IUPAC Gold book as:

$$c_\mathrm{solute}=\frac{n_\mathrm{solute}}{V_\mathrm{solution}}$$

Practical procedure is:

  1. Add solvent to the solute for the total volume to be little less than the target volume.
  2. Dissolve the solute
    • There is frequent volume contraction, $V_\text{solution} \lt V_\text{solute} + V_\text{solvent}$.
  3. Be sure no liquid drops are on the flask walls or the plug.
  4. Add the rest of solvent to reach the target volume and make the final mixing.

For precise preparation, this is being done in calibrated conical glass flasks with the narrow neck, marked for the target water level.

Your latter assumed procedure incorrectly assumes the solute density $\pu{1 g/mL}$ and that there is no volume change during dissolution.

Obviously, the procedure may not be that strict if accuracy requirements are low.

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@ Pono. Your two propositions are both wrong. You must introduce $11.76$ g solute in a volumetric flask having a volume of $400$ mL (which is not easy to find by the way). Then you add $300$ or $350$ mL water in the flask, and stir until the product is totally dissolved. Then you add enough water to get a final volume equal to exactly $400$ mL. This volume of water is not important to know, but the total amount of added water is probably different from $400$ mL and from $388.24$ mL. Remember the volumes are not additive in chemistry. $a$ mL solute (powder) plus $b$ mL solvant never gives a solution volume equal to $a + b$.

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