Question

A detergent $\left(C_{14}{H}_{29}{SO}_4^{-}{Na}^{+}\right)$ solution becomes a colloidal sol. at a concentration of ${10}^3$M on an average ${10}^{15}$ colloidal particles are present in ${100mm}^3$. What is the average number of ions are contains in one colloidal particle(micelle):$(N_A=6\times {10}^{23})$

Answer 1

${10}^{3}M\to {10}^3$ mol in $1L$

$\therefore$In $100m{m}^3\to ({10}^3\times {10}^2\times {10}^{-6})$

ions $={10}^{-1}{N}_A$

$=6\times {10}^{22}$

Avg ${10}^{15}$ particles

$\therefore$ No. of ions in 1 particle $=\frac{6\times {10}^{22}}{{10}^{15}}$

$=6\times {10}^7$