Final moles: \(pV = nRT\Rightarrow n_f = \frac{(11+1)\times10^5\times0.030}{\frac{100}{12}\times300}=\frac{12\times10^5\times0.030}{2500}=\frac{36000}{2500}=14.4\ \text{mol}\). Moles withdrawn \(= 18.20-14.4=3.8\ \text{mol}\). Note: absolute pressure \(= 11+1=12\ \text{atm}\). Mass \(= 3.8\times32=121.6\ \text{g}\approx 0.125\ \text{kg}\). Closest: option (3) 0.125 kg... but re-checking: \(n_f=\frac{12\times1.01\times10^5\times0.030}{(100/12)\times300}=\frac{3.636\times10^4}{2500}=14.54\). Withdrawn \(=3.66\ \text{mol}\times32=117\ \text{g}\approx0.116\ \text{kg}\). Answer: option (1).