Number of atoms in 558.5g $Fe$ (at.wt =55.85) is

Question

$\begin{array}{1 1}(a)\;Twice\;that\;in\;60g\;carbon\\(b)\;6.023\times 10^{22}\\(c)\;Half\;of\;8g\;He\\(d)\;558.5\times 6.023\times 10^{23}\end{array}$