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

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

### Related Formula Standard molecular weight calculation formula system: textMolar Mass = sum (textAtom Count times textAtomic Weight)$$\text{Molar Mass} = \sum (\text{Atom Count} \times \text{Atomic Weight})$$ textMass = textMoles times textMolar Mass$$\text{Mass} = \text{Moles} \times \text{Molar Mass}$$ ### Core Logic Let's tabulate and sum up the constituent atoms based on the verified structural formula layout : * Counting positions meticulously yields a molecular identity formula of mathrmC_9H_10FIN_2O_4$\mathrm{C_9H_{10}FIN_2O_4}$. * Calculating the total molar mass: textMass = (9 times 12) + (10 times 1) + 19 + 127 + (2 times 14) + (4 times 16)$$\text{Mass} = (9 \times 12) + (10 \times 1) + 19 + 127 + (2 \times 14) + (4 \times 16)$$ textMass = 108 + 10 + 19 + 127 + 28 + 64 = 372mathrm~g/mol$$\text{Mass} = 108 + 10 + 19 + 127 + 28 + 64 = 372\mathrm{~g/mol}$$

### Step 1: Final Mass Scaling Find the mass of 0.1mathrm~mol$0.1\mathrm{~mol}$ of the sample: textWeight = 0.1 times 372 = 37.2mathrm~g = 372 times 10^-1mathrm~g$$\text{Weight} = 0.1 \times 372 = 37.2\mathrm{~g} = 372 \times 10^{-1}\mathrm{~g}$$ Hence, the integer value for the blank box is 372. ### Pattern Recognition When counting atoms in nucleoside structures, remember that each vertex in the ribose sugar ring represents a fully saturated carbon component unless double bonds or heteroatoms are explicitly drawn. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules Class 11 Chemistry: Some Basic Concepts of Chemistry