Answer
702 g
Work Step by Step
We find:
$V=18.5\,L$
$P=11.2\,atm$
$T=(28.2+273)\,K=301.2\,K$
$n=\frac{PV}{RT}=\frac{11.2\,atm\times18.5\,L}{0.08206\,L\,atm\,K^{-1}mol^{-1}\times301.2\,K}$
$= 8.383\,mol$
Grams of Kr(g)= $8.383\,mol\,Kr \times\frac{83.798\,g}{1\,mol\,Kr}$
$= 702\,g$
