Question

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

Hint:

When calculating the weight of a compound, simply multiply the molar mass by the number of moles to get the total mass in grams.

Answer 1

Correct Answer: 372

The problem asks for the weight of 0.1 mol of the given antiviral compound (P). To find this, we first need to calculate the molar mass of the compound from its chemical structure and the provided atomic masses.

Concept Used:

The relationship between mass, number of moles, and molar mass is given by the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar Mass} \]

The molar mass of a compound is the sum of the atomic masses of all the atoms in its molecular formula.

Step-by-Step Solution:

Step 1: Determine the molecular formula of the compound (P) by counting the number of atoms of each element from its structure.

The compound consists of a 5-iodouracil base and a fluorinated deoxyribose sugar.

• Carbon (C): There are 4 carbon atoms in the pyrimidine ring and 5 carbon atoms in the sugar moiety. Total C atoms = 4 + 5 = 9.
• Hydrogen (H): There is 1 H on the pyrimidine ring (at C6), 1 H on a ring nitrogen (at N3), and 8 H atoms on the sugar moiety (at C1', C2', C3', C4', C5', and in the two -OH groups). Total H atoms = 1 + 1 + 8 = 10.
• Nitrogen (N): There are 2 nitrogen atoms in the pyrimidine ring.
• Oxygen (O): There are 2 oxygen atoms in the pyrimidine ring (as carbonyl groups) and 3 oxygen atoms in the sugar moiety (one in the furanose ring and two in -OH groups). Total O atoms = 2 + 3 = 5.
• Fluorine (F): There is 1 fluorine atom on the sugar moiety.
• Iodine (I): There is 1 iodine atom on the pyrimidine ring.

The molecular formula of compound (P) is \( \text{C}_9\text{H}_{10}\text{FIN}_2\text{O}_5 \).

Step 2: Calculate the molar mass of compound (P) using the given atomic masses.

Atomic masses (g/mol): H = 1, C = 12, N = 14, O = 16, F = 19, I = 127.

\[ \text{Molar Mass} = (9 \times \text{C}) + (10 \times \text{H}) + (1 \times \text{F}) + (1 \times \text{I}) + (2 \times \text{N}) + (5 \times \text{O}) \] \[ \text{Molar Mass} = (9 \times 12) + (10 \times 1) + (1 \times 19) + (1 \times 127) + (2 \times 14) + (5 \times 16) \] \[ \text{Molar Mass} = 108 + 10 + 19 + 127 + 28 + 80 \] \[ \text{Molar Mass} = 372 \, \text{g/mol} \]

Step 3: Calculate the mass of 0.1 mol of compound (P).

\[ \text{Mass} = \text{Number of moles} \times \text{Molar Mass} \] \[ \text{Mass} = 0.1 \, \text{mol} \times 372 \, \text{g/mol} = 37.2 \, \text{g} \]

Step 4: Express the result in the required format of \( \text{___} \times 10^{-1} \, \text{g} \).

We need to find a value \( x \) such that \( x \times 10^{-1} = 37.2 \).

\[ x = \frac{37.2}{10^{-1}} = 37.2 \times 10 = 372 \]

So, the mass is \( 372 \times 10^{-1} \, \text{g} \).

The value to be filled in the blank is 372.

Answer 2

Molar mass is given as 372 g/mol for compound (P). 
Hence, for 0.1 mole, the mass will be: \[ \text{Mass} = \text{Molar mass} \times \text{Number of moles} = 372 \times 0.1 = 37.2 \, \text{g} \]