To find the average number of ions contained by one colloidal particle (micelle) of the detergent solution, we can follow these steps:
### Step 1: Determine the volume of the solution
We know that the concentration of the detergent solution is \(10^{-3} \, \text{M}\) (moles per liter). Since we are dealing with \(1 \, \text{mm}^3\) of solution, we need to convert this volume to liters.
\[
1 \, \text{mm}^3 = 1 \times 10^{-3} \, \text{cm}^3 = 1 \times 10^{-6} \, \text{L}
\]
### Step 2: Calculate the number of moles of detergent in \(1 \, \text{mm}^3\)
Using the concentration and the volume, we can find the number of moles of detergent in \(1 \, \text{mm}^3\):
\[
\text{Number of moles} = \text{Concentration} \times \text{Volume} = 10^{-3} \, \text{mol/L} \times 1 \times 10^{-6} \, \text{L} = 10^{-9} \, \text{mol}
\]
### Step 3: Convert moles of detergent to molecules
Using Avogadro's number (\(6.022 \times 10^{23} \, \text{molecules/mol}\)), we can find the total number of detergent molecules present in \(1 \, \text{mm}^3\):
\[
\text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} = 10^{-9} \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol}
\]
Calculating this gives:
\[
\text{Number of molecules} = 6.022 \times 10^{14} \, \text{molecules}
\]
### Step 4: Determine the average number of colloidal particles
We are given that there are \(10^{13}\) colloidal particles in \(1 \, \text{mm}^3\).
### Step 5: Calculate the average number of molecules (ions) per colloidal particle
To find the average number of detergent molecules (ions) per colloidal particle, we divide the total number of molecules by the number of colloidal particles:
\[
\text{Average number of ions per particle} = \frac{\text{Total number of molecules}}{\text{Number of colloidal particles}} = \frac{6.022 \times 10^{14}}{10^{13}}
\]
Calculating this gives:
\[
\text{Average number of ions per particle} = 60.22 \approx 60
\]
### Final Answer
The average number of ions contained by one colloidal particle (micelle) is approximately **60**.
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