The mass of sodium acetate $$(CH_3COONa)$$ required to prepare $$250 \text{ mL}$$ of $$0.35 \text{ M}$$ aqueous solution is _____ g. (Molar mass of $$CH_3COONa$$ is $$82.02 \text{ g mol}^{-1}$$) Round off to the nearest integer.

We need to find the mass of sodium acetate required to prepare a given solution.

The formula for mass from molarity is:

$$ m = M \times V \times M_w $$

where $$M$$ is molarity, $$V$$ is volume in litres, and $$M_w$$ is molar mass.

Given:

$$M = 0.35$$ M, $$V = 250$$ mL $$= 0.250$$ L, $$M_w = 82.02$$ g/mol

Substituting:

$$ m = 0.35 \times 0.250 \times 82.02 = 0.0875 \times 82.02 = 7.177 \text{ g} $$

Rounding to the nearest integer: $$m \approx 7$$ g.

Therefore, the answer is $$\boxed{7}$$.