How many atoms of nitrogen are present in 218 g of dinitrogen pentoxide (N2O5)?

• A. 6.02 × 10^{23} atoms
• B. 1.44 × 10^{24} atoms
• C. 2.89 × 10^{23} atoms
• D. 3.01 × 10^{24} atoms

Answer: (B)

Verified step by step guidance

1. Calculate the molar mass of dinitrogen pentoxide (N\_2O\_5) by summing the atomic masses of its constituent atoms: 2 nitrogen atoms and 5 oxygen atoms. Use the atomic masses from the periodic table (approximately 14.01 g/mol for N and 16.00 g/mol for O). The molar mass is given by \(\text{Molar mass} = 2 \times 14.01 + 5 \times 16.00\) g/mol.
2. Determine the number of moles of N\_2O\_5 in 218 g by dividing the given mass by the molar mass calculated in step 1. Use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
3. Calculate the number of molecules of N\_2O\_5 by multiplying the number of moles by Avogadro's number, \$6.02 \times 10^{23}\( molecules/mol. Use the formula \)\text{molecules} = \text{moles} \times 6.02 \times 10^{23}$.
4. Since each molecule of N\_2O\_5 contains 2 nitrogen atoms, multiply the number of molecules by 2 to find the total number of nitrogen atoms. Use the formula \(\text{nitrogen atoms} = 2 \times \text{molecules}\).
5. Express the final answer in scientific notation to match the format of the given options.