Answer
$71.2amu$
Work Step by Step
From ideal gas law
$n=\frac{PV}{RT}$
We plug in the known values to obtain:
$n=\frac{\frac{755}{760}\times 0.273}{0.08206\times 373}=0.007692mol$
Now we can find the molar mass as
$Molar \space mass =\frac{0.548g}{0.007692mol}=71.2\frac{g}{mol}$
Hence the molecular weight is $71.2amu$