A 10^{-3} M solution of detergent (C_{12}H_{ | vedclass

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(C) First,convert the volume to liters: $1 \ mm^3 = (10^{-1} \ cm)^3 = 10^{-3} \ cm^3 = 10^{-3} \ mL = 10^{-6} \ L$. Calculate the total number of det

Vedclass · Accepted Answer

$60$

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(C) First,convert the volume to liters: $1 \ mm^3 = (10^{-1} \ cm)^3 = 10^{-3} \ cm^3 = 10^{-3} \ mL = 10^{-6} \ L$. Calculate the total number of detergent molecules in $1 \ mm^3$: $	ext{Number of molecules} = 	ext{Molarity} 	imes 	ext{Volume in L} 	imes N_A$ $= 10^{-3} \ mol/L 	imes 10^{-6} \ L 	imes 6 	imes 10^{23} \ mol^{-1} = 6 	imes 10^{14}$ molecules. Given that there are $10^{13}$ micelles in this volume,the number of detergent molecules per micelle is: $	ext{Number of ions per micelle} = \frac{	ext{Total molecules}}{	ext{Number of micelles}} = \frac{6 	imes 10^{14}}{10^{13}} = 60$.

$A$ $10^{-3} \ M$ solution of detergent $(C_{12}H_{25}SO_4^-Na^+)$ forms a colloidal sol. In $1 \ mm^3$ of the solution,there are on average $10^{13}$ colloidal particles. How many ions are present in one colloidal particle (micelle)? $(N_A = 6 \times 10^{23})$

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