Answer
(a) 274 L
(b) 7.71 L
(c) 0.0152 L
(d) 1340 L
Work Step by Step
Recall that at STP, one mole of a gas occupies 22.4 L volume.
(a) Moles of He= $48.9\,g\,He\times\frac{1\,mol\,He}{4.00\,g\,He}$
$= 12.225\,mol\,He$
$V=12.225\times22.4\,L= 274\,L$
(b) Moles of Xe= $45.2\,g\,Xe\times\frac{1\,mol\,Xe}{131.293\,g\,Xe}$
$=0.34427\,mol\,Xe$
$V=0.34427\times22.4\,L= 7.71\,L$
(c) V= $48.2\times10^{-3}\,g\,Cl_{2}\times\frac{1\,mol\,Cl_{2}}{70.906\,g\,Cl_{2}}\times\frac{22.4\,L}{1\,mol\,Cl_{2}}$
$= 0.0152\,L$
(d) V= $3.83\times10^{3}\,g\,SO_{2}\times\frac{1\,mol\,SO_{2}}{64.066\,g\,SO_{2}}\times\frac{22.4\,L}{1\,mol\,SO_{2}}$
$=1340\,L$
