What is the mass in grams of 1.02 × 10^{24} atoms of manganese (Mn)? (Atomic mass of Mn = 54.94 g/mol)

• A. 54.9 g
• B. 9.3 g
• C. 93.1 g
• D. 16.2 g

Answer: (C)

Verified step by step guidance

1. Identify the given information: the number of atoms of manganese (Mn) is \$1.02 \times 10^{24}$ atoms, and the atomic mass of Mn is 54.94 g/mol.
2. Recall that 1 mole of any element contains Avogadro's number of atoms, which is \$6.022 \times 10^{23}$ atoms/mol.
3. Calculate the number of moles of Mn atoms by dividing the given number of atoms by Avogadro's number using the formula: \(\text{moles} = \frac{\text{number of atoms}}{6.022 \times 10^{23}}\).
4. Convert moles of Mn to mass in grams by multiplying the number of moles by the atomic mass of Mn: \(\text{mass (g)} = \text{moles} \times 54.94 \, \text{g/mol}\).
5. Combine the steps to find the mass of Mn corresponding to \$1.02 \times 10^{24}$ atoms without calculating the final value.