Question

what is the mass in grams of 1.02×10²⁴ atoms manganese (mn)?
0.169×10¹ g
0.112×10¹ g
9.30×10¹ g
9.30×10⁻¹ g

Explanation:

Step1: Recall Avogadro's number and molar mass

Avogadro's number is \(6.022\times 10^{23}\) atoms/mol, and the molar mass of Mn (manganese) is approximately \(54.94\) g/mol. We need to convert the number of atoms to moles first, then to grams.

Step2: Calculate moles of Mn

Use the formula \(n=\frac{N}{N_A}\), where \(N = 1.02\times 10^{24}\) atoms and \(N_A=6.022\times 10^{23}\) atoms/mol.
\[n=\frac{1.02\times 10^{24}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}\approx1.694\text{ mol}\]

Step3: Calculate mass of Mn

Use the formula \(m = n\times M\), where \(n\) is moles and \(M = 54.94\) g/mol.
\[m=1.694\text{ mol}\times54.94\text{ g/mol}\approx93.0\text{ g}=9.30\times 10^{1}\text{ g}\]

Answer:

\(9.30\times 10^{1}\text{ g}\) (corresponding to the option \(9.30\times 10^{1}\text{ g}\))