Problem 36 - A Complete Resource Book in Chemistry for JEE Main

Question

The weight of one molecule of a compound \(\mathrm{C}_{60} \mathrm{H}_{122}\) is (a) \(1.3 \times 10^{-20} \mathrm{~g}\) (b) \(5.01 \times 10^{-21} \mathrm{~g}\) (c) \(3.72 \times 10^{23} \mathrm{~g}\) (d) \(1.4 \times 10^{-21} \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(1.4 \times 10^{-21} \mathrm{~g}\).

Step 1: Determine Molar Mass of the Compound

The compound is \(\mathrm{C}_{60} \mathrm{H}_{122}\). First, calculate the molar mass. Carbon \((C)\) has a molar mass of approximately \(12.01 \, \mathrm{g/mol}\), and Hydrogen \((H)\) has a molar mass of approximately \(1.01 \, \mathrm{g/mol}\). Thus, the molar mass of \(\mathrm{C}_{60} \mathrm{H}_{122}\) is calculated as:\[60 \times 12.01 \, \mathrm{g/mol} + 122 \times 1.01 \, \mathrm{g/mol} = 720.6 + 123.22 = 843.82 \, \mathrm{g/mol}\]

Step 2: Use Avogadro's Number for Molecular Weight

Avogadro's number \(6.022 \times 10^{23}\) molecules/mol allows us to find the weight of a single molecule. Divide the molar mass by Avogadro's number to determine the weight of a single molecule:\[\frac{843.82 \, \mathrm{g/mol}}{6.022 \times 10^{23} \, \mathrm{molecules/mol}} = 1.4 \times 10^{-21} \, \mathrm{g/molecule}\]

Step 3: Compare with Given Options

The calculated weight of one molecule \(1.4 \times 10^{-21} \, \mathrm{g/molecule}\) matches option (d). Therefore, the correct answer is (d).

Key Concepts

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry, representing the number of atoms, ions, or molecules in one mole of a substance. A mole is a standard unit in chemistry that quantifies the amount of substance. Avogadro's number is approximately equal to

• \(6.022 \times 10^{23}\) molecules/mol.

This vast number helps bridge the gap between the atomic scale and day-to-day scales. When dealing with a single molecule, such as in the compound \(\text{C}_{60} \text{H}_{122}\), Avogadro's number allows us to convert the molar mass of the compound to the mass of a single molecule. This conversion is crucial because it gives scientists the ability to relate macroscopic measurements to the microscopic scale of molecules. By dividing the molar mass of a compound by Avogadro's number, we can find the weight of one molecule in grams. In our exercise, this number helped us determine that the weight of one molecule of \(\text{C}_{60} \text{H}_{122}\) is \(1.4 \times 10^{-21} \ \text{g}\). This process showcases the application of Avogadro's number in calculating molecular weight permeating everyday chemical calculations.

Molecular Weight

Molecular weight is a measure of the mass of one molecule of a substance and directly correlates with its molar mass. It is typically expressed in atomic mass units (amu) or grams per mole. To calculate the molecular weight, we sum up the atomic weights of all atoms present in a molecule. For instance, in the compound \(\text{C}_{60} \text{H}_{122}\), knowing the atomic weights of carbon and hydrogen is essential.

• Carbon (C) has an atomic weight of approximately \(12.01 \, \text{g/mol}\).
• Hydrogen (H) has an atomic weight of around \(1.01 \, \text{g/mol}\).

To find the molar mass of this compound, we calculate:\[60 \times 12.01 \, \text{g/mol} + 122 \times 1.01 \, \text{g/mol} = 843.82 \, \text{g/mol}\]Knowing the molar mass allows us to use Avogadro's number to find the mass of a single molecule, linking the macroscopic measurement with molecular-level understanding. This is crucial for determining how substances interact and understanding chemical reactions at a molecular level.

Carbon and Hydrogen Composition

Understanding the composition of a molecule such as \(\text{C}_{60} \text{H}_{122}\) is important for calculating its molecular weight and analyzing its chemical properties. The compound consists of 60 carbon atoms and 122 hydrogen atoms. This specific arrangement defines not only the molecular weight but also greatly influences the compound's characteristics due to the unique properties of carbon and hydrogen.Carbon is a versatile element that forms the backbone of many organic compounds. Its ability to form four covalent bonds allows for complex molecular structures.

• Each carbon atom contributes \(12.01 \, \text{g/mol}\) to the compound's molar mass.

Hydrogen, while much lighter, plays a significant role in molecular chemistry. Its valency helps stabilize structures by forming bonds with other elements like carbon.

• Each hydrogen atom contributes \(1.01 \, \text{g/mol}\) to the molar mass.

The balance and interaction between carbon and hydrogen in this compound are crucial for defining its physical properties and reactivity, making understanding its composition fundamental to application and synthesis in chemistry.