Given the surtactrant solution has molar concentration is 10−3M . Volume ofsolution = 10mL ∴ no. of moles of surtactant =10×10−3×10−3 moles =10−5 moles No. of polarheads = 6×1023×10−5 =6×1018 Area coverd on polar substrate =0.24×10{4}m2 a2×6×1018=2.4×10−5 a2=4×10−24 a=2×10−12m ∴a=2.0pm