0.1 mol of the following given antiviral compound (P) will weigh _______ $$\times 10^{-1}$$ g.

(Given: molar mass in g mol$$^{-1}$$ H: 1, C: 12, N: 14, O: 16, F: 19, I: 127)

The molecular formula of the given antiviral compound is obtained by counting the atoms present in the heterocyclic base and the substituted sugar moiety.

For the nitrogenous base:

• Carbon atoms = 4
• Hydrogen atoms = 2
• Nitrogen atoms = 2
• Oxygen atoms = 2
• Iodine atoms = 1

Hence, the contribution is

$$\mathrm{C_4H_2N_2O_2I}$$

For the sugar moiety:

• Carbon atoms = 5
• Hydrogen atoms = 8
• Oxygen atoms = 3
• Fluorine atoms = 1

Hence, the contribution is

$$\mathrm{C_5H_8O_3F}$$

Therefore, the molecular formula of the compound is

$$\mathrm{C_9H_{10}N_2O_5FI}$$

Using the given atomic masses,

• Carbon: $$9 \times 12 = 108$$
• Hydrogen: $$10 \times 1 = 10$$
• Nitrogen: $$2 \times 14 = 28$$
• Oxygen: $$5 \times 16 = 80$$
• Fluorine: $$1 \times 19 = 19$$
• Iodine: $$1 \times 127 = 127$$

Hence, the molar mass is

$$108 + 10 + 28 + 80 + 19 + 127 = 372\ \mathrm{g\ mol^{-1}}$$

The mass corresponding to $$0.1$$ mole is

$$\text{Mass} = 0.1 \times 372 = 37.2\ \mathrm{g}$$

Expressing this in the required form,

$$37.2\ \mathrm{g} = 372 \times 10^{-1}\ \mathrm{g}$$

Therefore, the required integer value is

$$\boxed{372}$$