Question

determine the mass in grams of 4.93e21 atoms of barium. the mass of one mole of barium is 137.33 g.

Explanation:

Step1: Find moles of barium

Use Avogadro's number ($N_A = 6.022\times10^{23}$ atoms/mol). Moles ($n$) = $\frac{\text{number of atoms}}{N_A}$ = $\frac{4.93\times10^{21}}{6.022\times10^{23}}$

Step2: Calculate mass

Mass ($m$) = moles × molar mass. Molar mass = 137.33 g/mol. So $m = \frac{4.93\times10^{21}}{6.022\times10^{23}} \times 137.33$
First, $\frac{4.93\times10^{21}}{6.022\times10^{23}} \approx 0.00819$ mol. Then $0.00819 \times 137.33 \approx 1.125$ g.

Answer:

1.13 (or ~1.12 - 1.13, depending on precision)