The correct option is D 0.526 L
Total moles of NH3 adsorbed on surface =1.415×10226.023×1023=0.0235.
Moles of NH3 before adsorption =100×21000=0.2.
So, moles of NH3 left after adosrption = moles of HCl required =0.2−0.0235=0.175.
Also, moles of HCl required to clean surface = moles of NH3 adsorbed.
nHCl=0.0235
∴1×V=0.0235×0.0821×273
VHCl(g)=0.526 L