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 $\pu{400 ml}?$