Answer
b) 4.77 g/L is correct answer
Work Step by Step
Given :- molar mass = 146.07 g/mol , pressure (P) = 1.00 atm , Temperature (T) = 100$^{\circ}$C = 373 K , The gas constant (R) = 0.08206 L atm $k^{-1} mol^{-1}$
Now, we have to determine the density of the gas
we know the ideal gas equation i;e
PV=nRT on further evaluation this we get the relation with density and molar mass
PM=dRT
density (d) = $\frac{PM}{RT}$=$\frac{1.00 atm \times 146.07g/mol}{0.08206 L \space atm \space k^{-1} mol^{-1} \times \space 373 K}$ = 4.7722 $\approx$ 4.77 g/L
The density of the gas is 4.77 g/L.
Therefore option b) is right.
