Answer
0.90 g
Work Step by Step
Pressure $P=1.1\,atm$
Volume $V=5.0\,L$
Universal gas constant $R= 0.0821\,L\,atm\,mol^{-1}K^{-1}$
Temperature $T=(25\,+273)K= 298\,K$
Recall that $PV=nRT$ (ideal gas law)
$\implies $ moles of helium $n= \frac{PV}{RT}$
$=\frac{(1.1\,atm)(5.0\,L)}{(0.0821\,L\,atm\,mol^{-1}K^{-1})(298\,K)}$
$=0.225\,mol$
Mass of helium required=
$n\times$ molar mass of helium
$=0.225\,mol\times 4.00\,g/mol=0.90\,g$
