Q. At a concentration of $10^{- 3}M$ , a detergent $\left(\left(\text{C}\right)_{12} \left(\text{H}\right)_{25} \text{SO}_{4}^{-} \left(\text{Na}\right)^{+}\right)$ solution becomes a colloidal solution. On an average $10^{13}$ colloidal particles are present in $1mm^{3}$ . Calculate the average number of ions contained in one colloidal particle (micelle)? [ Given $N_{A}=6\times 10^{23}$ ]

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$6\times 10^{7}$

$10$

$60$

None of these

Solution:

No. of sodium lauryl sulphate $\left(\text{CH}\right)_{3} \left(\left(\text{CH}\right)_{2}\right)_{11} \text{SO}_{4}^{-} \left(\text{Na}\right)^{+}\right)$ in $1$ litre solution
= $10^{- 3}\times 6\times 10^{23}$
= $6\times 10^{20}$
No. of sodium lauryl sulphate per $mm^{3}$ = $6\times 10^{14}$
No. of colloidal particles per $mm^{3}$ = $10^{13}$
No. of molecules per colloidal particle
$= \frac{6 \times 1 0^{1 4}}{1 0^{1 3}}$
= $60$

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