Question

One mole of glucose molecules contains approximately:

• A. 3.01 $\times$ 10$^{23}$ molecules
• B. 6.02 $\times$ 10$^{23}$ molecules
• C. 1.204 $\times$ 10$^{24}$ molecules
• D. 9.03 $\times$ 10$^{23}$ molecules

Hint:

Remember that the value of Avogadro's number ($N_A = 6.022 \times 10^{23}$) is independent of the identity of the chemical species.
Whether you are dealing with one mole of glucose, one mole of water, or one mole of electrons, the count of those fundamental entities will always be $6.02 \times 10^{23}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question is from the topic "Some Basic Concepts of Chemistry," specifically focusing on the Mole Concept.
Our objective is to identify the number of molecules present in a single mole of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$).

Step 2: Key Formula or Approach:
We use the definition of the mole and Avogadro's constant.
The formula relating the number of moles ($n$) to the number of molecules ($N$) is:
\[ N = n \times N_A \]
where $N_A$ is Avogadro's number, approximately equal to $6.022 \times 10^{23} \text{ molecules/mol}$.

Step 3: Detailed Explanation:

• A mole is defined as the amount of a substance that contains exactly the same number of fundamental entities (such as atoms, molecules, or ions) as there are atoms in $12 \text{ grams}$ of pure Carbon-12.

• This constant value is designated as Avogadro's number, $N_A \approx 6.022 \times 10^{23}$.

• Glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) is a molecular compound, which means that its basic constituent units are molecules.

• Therefore, exactly one mole of glucose contains $1 \times N_A$ glucose molecules, which is approximately $6.02 \times 10^{23}$ molecules.

• If we were to calculate the number of individual atoms, one mole of glucose contains $6$ moles of Carbon atoms, $12$ moles of Hydrogen atoms, and $6$ moles of Oxygen atoms, totaling $24 \times 6.02 \times 10^{23}$ atoms.



Step 4: Final Answer:
Thus, one mole of glucose molecules contains approximately $6.02 \times 10^{23}$ molecules, matching option (B).