# 57.7 g Ni contains how many atoms?

$\text{Number of atom}$ $=$ $\text{Mass"/"Molar mass"xx"Avogadro's Number.}$
And thus, $\frac{57.7 \cdot g}{58.7 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$
$\cong 6.022 \times {10}^{23}$ $\text{nickel atoms}$.