Answer
$3.2\times 10^2amu$
Work Step by Step
From ideal gas law
$n=\frac{PV}{RT}$
We plug in the known values to obtain:
$n=\frac{\frac{786}{760}\times 0.250}{0.08206\times 394}=0.007996mol$
Now we can find the molar mass as
$Molar \space mass =\frac{2.56g}{0.007996mol}=3.20\times 10^2\frac{g}{mol}$
Hence the molecular weight is $3.2\times 10^2amu$
