[Solution by Claude - not from source] Absolute pressure $=11+1=12$ atm $=1.212\times10^6$ Pa. $n_f=\frac{PV}{RT}=\frac{1.212\times10^6\times30\times10^{-3}}{(100/12)\times300}\approx14.54$ mol. Withdrawn $=18.20-14.54\approx3.66$ mol $\approx0.117$ kg, closest to 0.116 kg. Answer (3).