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

Concept: Number of particles in one mole is given by Avogadro's number. \[ N_A = 6.022 \times 10^{23} \] Each molecule of glucose \(C_6H_{12}O_6\) contains \(6\) carbon atoms.

Step 1: Find the number of glucose molecules. \[ 0.35 \times 6.022 \times 10^{23} \] \[ = 2.1077 \times 10^{23} \text{ molecules} \]

Step 2: Multiply by carbon atoms per molecule. \[ \text{Carbon atoms} = 6 \times 2.1077 \times 10^{23} \] \[ = 1.2646 \times 10^{24} \] \[ \approx 1.26 \times 10^{24} \] Thus, the number of carbon atoms present is: \[ 1.26 \times 10^{24} \]