Answer:
6.3 × 10²⁵ atoms of O
Explanation:
[tex]\text{moles of O} = \text{9.4 mol C$_{12}$H$_{22}$O$_{11}$} \times \frac{ 11 \text{ mol O }}{\text{1 mol C$_{12}$H$_{22}$O$_{11}$}} = \text{103.4 mol O}[/tex]
[tex]\text{Atoms of O} =\text{103.4 mol O} \times \frac{ 6.022 \times 10^{23} \text{ atoms O}}{\text{1 mol O}}= 6.3 \times 10^{25} \text{ atoms O}[/tex]
Thus, 9.4 mol of C₁₂H₂₂O₁₁ contains 6.3 × 10²⁵ atoms of O.
