Excess of NaOH (aq) was added to 100 mL of FeCl$$_3$$ (aq) resulting into 2.14 g of Fe(OH)$$_3$$. The molarity of FeCl$$_3$$(aq) is: (Given the molar mass of Fe = 56 g mol$$^{-1}$$ and molar mass of Cl = 35.5 g mol$$^{-1}$$)

We begin by writing the balanced chemical reaction between ferric chloride and an excess of sodium hydroxide: $$FeCl_3 \;+\; 3\,NaOH \;\rightarrow\; Fe(OH)_3 \;+\; 3\,NaCl$$. From this equation we see that one mole of $$FeCl_3$$ produces one mole of the precipitate $$Fe(OH)_3$$.

The mass of the precipitate obtained in the experiment is given as 2.14 g. To convert this mass into moles, we first need the molar mass of $$Fe(OH)_3$$. We list every atomic contribution:

$$\text{Molar mass of }Fe(OH)_3 = M_{Fe} + 3\,(M_O + M_H).$$

Substituting the atomic molar masses $$M_{Fe}=56\;g\,mol^{-1},\; M_O=16\;g\,mol^{-1},\; M_H=1\;g\,mol^{-1}$$, we obtain

$$ \begin{aligned} M_{Fe(OH)_3} &= 56 + 3\,(16+1)\\ &= 56 + 3\times17\\ &= 56 + 51\\ &= 107\;g\,mol^{-1}. \end{aligned} $$

Now we calculate the moles of $$Fe(OH)_3$$ precipitated using the definition $$\text{moles} = \dfrac{\text{mass}}{\text{molar mass}}$$:

$$ n_{Fe(OH)_3} = \dfrac{2.14\;g}{107\;g\,mol^{-1}} = 0.0200\;mol. $$

Because the stoichiometric ratio between $$FeCl_3$$ and $$Fe(OH)_3$$ is 1 : 1, the moles of $$FeCl_3$$ originally present are also $$0.0200\;mol$$.

The volume of the $$FeCl_3$$ solution used is 100 mL, which we convert to litres for molarity calculations: $$100\;mL = 0.100\;L$$.

Molarity is defined as $$M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$$. Substituting the values we have just obtained gives

$$ M_{FeCl_3} = \dfrac{0.0200\;mol}{0.100\;L} = 0.200\;M. $$

Hence, the correct answer is Option B.