Chemistry - Surface Chemistry Question with Solution | TestHub

At$\mathrm{C}\mathrm{M}\mathrm{C},\mathrm{}$the number of cationic colloidal particles$={10}^{-3}\times 6.0\times {10}^{23}$$\mathrm{=\; 6.0}\times {10}^{20}\mathrm{}$per L

Therefore, number of particles per$\mathrm{m}{\mathrm{m}}^3$
$= \frac{6.0\times {10}^{20}}{{10}^6}\left(\because 1\text{L}={10}^6{\text{mm}}^3\right)$
$= 6.0\times {10}^{14}$particles/$\mathrm{m}{\mathrm{m}}^3$

Number of micelles in$\mathrm{1 }\mathrm{}\mathrm{m}{\mathrm{m}}^3{\text{= 10}}^{13}$(given)
Number of cations forming one micelle
$=\frac{6.0\times {10}^{14}}{{10}^{13}}$
$=60$cations