# Question #a292a

May 14, 2017

$113 g$
Start with the given quantity $1.70 \times {10}^{25}$ atoms $H e$, and recognize that there are $4.003 g H e$ in $6.02 \times {10}^{23}$ atoms $H e$:
$g H e = 1.70 \times {10}^{25} a t o m s H e \frac{4.003 g H e}{6.02 \times {10}^{23} a t o m s H e} = 113 g H e$