$n_{Fe_2(SO_4)_3}=\dfrac{4}{400}=0.01(mol)$

$C_M=\dfrac{0.01}{0.1}=0.1(M)$

\[
\begin{array}{c|c c c}
 PTPL& Fe_2(SO_4)_3 & \rightarrow &2Fe^{3+} &+3SO_4^{2-} \\
\hline
 \text{Phân ly}&  0.1 &  &\rightarrow 0.2 & \rightarrow 0.3 (M)\\

\end{array}
\]

$C_M(Fe^{3+})=\dfrac{0.2}{0.1}=2(M)$