What mass of the compound $\ce{CrO3}$ $\text{(M = 100.0)}$ contains $4.5\times10^{23}$ oxygen atoms?

$ \textbf{(A) }\text{2.25 g}\qquad\textbf{(B) }\text{12.0 g}\qquad\textbf{(C) }\text{25.0 g}\qquad\textbf{(D) }\text{75.0 g}\qquad$
1. First, we need to determine the number of moles of oxygen atoms in $4.5 \times 10^{23}$ atoms. Using Avogadro's number ($6.022 \times 10^{23}$ atoms/mol), we can convert the number of atoms to moles:
   \[
   \text{moles of oxygen atoms} = \frac{4.5 \times 10^{23} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}} = \frac{4.5}{6.022} \text{ mol}
   \]

2. Next, we need to determine the number of moles of $\ce{CrO3}$ that contain this number of moles of oxygen atoms. Each molecule of $\ce{CrO3}$ contains 3 oxygen atoms. Therefore, the number of moles of $\ce{CrO3}$ is:
   \[
   \text{moles of } \ce{CrO3} = \frac{\text{moles of oxygen atoms}}{3} = \frac{\frac{4.5}{6.022}}{3} = \frac{4.5}{3 \times 6.022} \text{ mol}
   \]

3. Now, we need to find the mass of this amount of $\ce{CrO3}$. The molar mass of $\ce{CrO3}$ is given as 100.0 g/mol. Therefore, the mass of $\ce{CrO3}$ is:
   \[
   \text{mass of } \ce{CrO3} = \left( \frac{4.5}{3 \times 6.022} \text{ mol} \right) \times 100.0 \text{ g/mol}
   \]

4. Simplifying the expression:
   \[
   \text{mass of } \ce{CrO3} = \left( \frac{4.5}{18.066} \text{ mol} \right) \times 100.0 \text{ g/mol} = \left( \frac{4.5}{18.066} \right) \times 100.0 \text{ g}
   \]

5. Calculating the final value:
   \[
   \text{mass of } \ce{CrO3} = \left( 0.249 \text{ mol} \right) \times 100.0 \text{ g/mol} = 25.0 \text{ g}
   \]

Conclusion:
\[
\boxed{25.0 \text{ g}}
\]