From $$6.55$$ g of aniline, the maximum amount of acetanilide that can be prepared will be ______ $$\times 10^{-1}$$ g.

Aniline is first converted to acetanilide by acetylation. The balanced equation is

$$C_6H_5NH_2 + CH_3COOH \;(\text{or } (CH_3CO)_2O)\;\rightarrow\; C_6H_5NHCOCH_3 + H_2O$$

One mole of aniline gives one mole of acetanilide, so the reaction follows a 1 : 1 molar ratio.

Step 1 - Molar mass of aniline
$$M_{\text{aniline}} = 6(12) + 7(1) + 14 = 72 + 7 + 14 = 93 \text{ g mol}^{-1}$$

Step 2 - Moles of aniline taken
Given mass of aniline = $$6.55\text{ g}$$
$$n_{\text{aniline}} = \frac{6.55}{93} = 0.07043 \text{ mol}$$ (keep four significant figures for accuracy)

Step 3 - Molar mass of acetanilide
Acetanilide formula: $$C_8H_9NO$$
$$M_{\text{acetanilide}} = 8(12) + 9(1) + 14 + 16 = 96 + 9 + 14 + 16 = 135 \text{ g mol}^{-1}$$

Step 4 - Maximum (theoretical) mass of acetanilide
Moles of acetanilide produced = moles of aniline (1 : 1 ratio): $$0.07043 \text{ mol}$$
$$m_{\text{acetanilide}} = 0.07043 \times 135 = 9.517 \text{ g}$$

Step 5 - Expressing the answer as “×10−1 g”
Write 9.517 g as a multiple of $$10^{-1}$$ g:
$$9.517 \text{ g} = 95.17 \times 10^{-1} \text{ g}$$

Rounded to the nearest whole number (as required in JEE integer-type answers), the coefficient is $$95$$.

Therefore, the maximum amount of acetanilide obtainable is $$\boxed{95 \times 10^{-1}\text{ g}}$$.