The addition of dilute NaOH to Cr$$^{3+}$$ salt solution will give:

First, remember the common precipitation rule for metal ions: a metal ion of charge $$n+$$ combines with $$n$$ hydroxide ions to give its hydroxide as a solid.

$$\text{M}^{\,n+}+n\,\text{OH}^- \;\longrightarrow\; \text{M(OH)}_n\downarrow$$

Applying this rule to chromium(III) ion, which carries a charge of $$3+$$, we write

$$\text{Cr}^{\,3+}+3\,\text{OH}^- \;\longrightarrow\; \text{Cr(OH)}_3\downarrow$$

Because the NaOH is only dilute, the solution supplies just enough $$\text{OH}^-$$ ions to reach the point of precipitation; there is no large excess present that could dissolve the precipitate and form the soluble complex $$[\text{Cr(OH)}_6]^{3-}$$. At this stage we therefore obtain a green, gelatinous solid.

Chromium(III) hydroxide is amphoteric and tends to polymerise by losing water. The solid actually separates in a hydrated, polymeric form that can be represented as hydrated chromium(III) oxide:

$$2\,\text{Cr}^{\,3+}+6\,\text{OH}^-+3\,\text{H}_2\text{O}\;\longrightarrow\;\text{Cr}_2\text{O}_3(\text{H}_2\text{O})_3\downarrow$$

Since the precise number of water molecules is variable, we usually write the product in the more general form $$\text{Cr}_2\text{O}_3(\text{H}_2\text{O})_n$$, emphasising that it is a hydrated oxide rather than the simple monomer $$\text{Cr(OH)}_3$$.

Thus, on adding only dilute NaOH to a solution of $$\text{Cr}^{\,3+}$$ ions, the immediate product is a precipitate of hydrated chromium(III) oxide.

Among the listed options, this matches “precipitate of $$\text{Cr}_2\text{O}_3(\text{H}_2\text{O})_n$$”.

Hence, the correct answer is Option C.