Answer
$47.7amu$
Work Step by Step
From ideal gas law
$n=\frac{PV}{RT}$
We plug in the known values to obtain:
$n=\frac{\frac{797}{760}\times 1}{0.08206\times 363}=0.033262mol$
Now we can find the molar mass as
$Molar \space mass =\frac{1.585g}{0.033262mol}=47.651\frac{g}{mol}$
Hence the molecular weight is $47.7amu$
