Question

Calculate the number of carbon atoms present in \(0.35\) mole of glucose \((C_6H_{12}O_6)\).

• A. \(1.26 \times 10^{23}\)
• B. \(1.26 \times 10^{24}\)
• C. \(2.10 \times 10^{24}\)
• D. \(3.50 \times 10^{23}\)

Hint:

To find atoms in a compound: \(\text{Moles} \times N_A \times \text{number of that atom in the formula}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the total number of individual carbon atoms in a given quantity (moles) of glucose molecules.
Step 2: Key Formula or Approach:
Number of atoms = Moles \(\times\) Avogadro's Number (\(N_A\)) \(\times\) Atomicity of Carbon.
\(N_A = 6.022 \times 10^{23}\).
Step 3: Detailed Explanation:
In 1 molecule of glucose (\(C_6H_{12}O_6\)), there are 6 Carbon atoms.
Total moles of Carbon = \(0.35\ \text{mol glucose} \times 6\ \text{atoms C/molecule} = 2.1\ \text{mol of Carbon atoms}\).
Total number of Carbon atoms = \(2.1 \times 6.022 \times 10^{23}\).
\[ \text{Atoms} = 12.64 \times 10^{23} = 1.26 \times 10^{24} \] Step 4: Final Answer:
The number of carbon atoms is \( 1.26 \times 10^{24} \).