How many atoms are present in 179.0 g of iridium (Ir)? (Atomic mass of Ir = 192.2 g/mol)

• A. 1.12 × 10^{24} atoms
• B. 9.36 × 10^{22} atoms
• C. 3.45 × 10^{24} atoms
• D. 5.61 × 10^{23} atoms

Answer: (D)

Verified step by step guidance

1. Identify the given information: mass of iridium (Ir) is 179.0 g and the atomic mass of Ir is 192.2 g/mol.
2. Calculate the number of moles of iridium using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{179.0\ \text{g}}{192.2\ \text{g/mol}}\).
3. Recall that one mole of any element contains Avogadro's number of atoms, which is \$6.022 \times 10^{23}$ atoms/mol.
4. Calculate the total number of atoms by multiplying the moles of iridium by Avogadro's number: \(\text{atoms} = \text{moles} \times 6.022 \times 10^{23}\).
5. Express the final answer in scientific notation to compare with the given options.