# Question #6de44

Molar mass of $H e$ = $4$ $g \cdot m o {l}^{-} 1$
So, $6.022 \cdot {10}^{23}$ atoms of $H e$ = $4$ $g$
So, $0.54$ $g$ of $H e$ = $\frac{6.022 \cdot {10}^{23} \cdot 0.54}{4}$ = $8.1297 \cdot {10}^{22}$ atoms