The critical micelle concentration (CMC) of a cationic colloidal electrolyte is `10^(-3)` M. If 1 `mm^(3)` contains `10^(13)` micelles, the number of cations making one micells is
(Given, `N_(A) = 6.0 xx 10^(23) mol^(-1)`)

Answer

To solve the problem step by step, we will follow the given information and apply the necessary formulas.

### Step 1: Understand the Given Information
- The critical micelle concentration (CMC) of the cationic colloidal electrolyte is \(10^{-3}\) M.
- The volume of the solution is \(1 \, \text{mm}^3\) which is equivalent to \(1 \times 10^{-3} \, \text{mL}\).
- The number of micelles in this volume is \(10^{13}\).
- Avogadro's number \(N_A = 6.0 \times 10^{23} \, \text{mol}^{-1}\).

### Step 2: Calculate the Number of Moles of Cationic Colloids
Using the definition of molarity:
\[ \text{Molarity (C)} = \frac{\text{moles of solute (n)}}{\text{volume of solution (L)}} \]
We can rearrange this to find the number of moles of cationic colloids:
\[ n = C \times V \]
Where:
- \(C = 10^{-3} \, \text{mol/L}\)
- \(V = 1 \, \text{mm}^3 = 1 \times 10^{-3} \, \text{mL} = 1 \times 10^{-6} \, \text{L}\)

Now substituting the values:
\[ n = 10^{-3} \, \text{mol/L} \times 1 \times 10^{-6} \, \text{L} = 10^{-9} \, \text{mol} \]

### Step 3: Calculate the Number of Cations in One Micelle
Since we have \(10^{13}\) micelles in \(1 \, \text{mm}^3\), we can find the number of moles of cationic colloids per micelle:
\[ \text{Moles of cationic colloids per micelle} = \frac{n}{\text{Number of micelles}} = \frac{10^{-9} \, \text{mol}}{10^{13}} = 10^{-22} \, \text{mol} \]

### Step 4: Convert Moles to Number of Cations
To find the number of cations in one micelle, we multiply the number of moles of cationic colloids per micelle by Avogadro's number:
\[ \text{Number of cations} = \text{Moles} \times N_A = 10^{-22} \, \text{mol} \times 6.0 \times 10^{23} \, \text{mol}^{-1} \]
Calculating this gives:
\[ \text{Number of cations} = 6.0 \times 10^{1} = 60 \]

### Final Answer
The number of cations making one micelle is **60**.

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