Answer
$1.9\times10^{6}\,g$
Work Step by Step
$V=1.2\times10^{7}\,L$
$P=737\,mmHg=737\,mmHg\times\frac{1\,atm}{760\,mmHg}$
$=0.9697\,atm$
$T=25^{\circ}C=(25+273)K=298\,K$
$R=0.0821\,L\,atm\,K^{-1}mol^{-1}$
Using ideal gas law, $PV=nRT$, we get
$n=\frac{PV}{RT}=\frac{0.9697\,atm\times1.2\times10^{7}\,L}{0.0821\,L\,atm\,K^{-1}mol^{-1}\times298\,K}=4.756\times10^{5}\,mol$
Mass in grams=$n\times\text{molar mass of He}$
$=4.756\times10^{5}\,mol\times4.00\,g/mol=1.9\times10^{6}\,g$
