JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Surface Chemistry

First count the surfactant molecules spread over the surface. The amount present is $\frac{1\times 10^{-3}\text{mol/L}\times 10\times 10^{-3}\text{L}}{1}=10^{-5}$ mol, which is $10^{-5}\times 6\times 10^{23}=6\times 10^{18}$ molecules. These cover $0.24{\text{cm}}^2$, so each head takes up $\frac{0.24}{6\times 10^{18}}=4\times 10^{-20}{\text{cm}}^2$. Treating a head as a cube, that area is one face $a^2$, so $a=\sqrt{4\times 10^{-20}}=2\times 10^{-10}\text{cm}=2\times 10^{-12}\text{m}$, that is $2$ pm.