# What is the mass of 1.39 mol of MgO?

$1$ $m o l$ of magnesium oxide has a mass of $40.30$ $g \cdot m o {l}^{-} 1$
So the required mass is $1.39$ $\cancel{m o l}$ $\times$ $40.30$ $g \cdot \cancel{m o {l}^{-} 1}$ $=$ ?? $g$.