To solve the problem of determining the edge length of a cube-shaped surfactant molecule adsorbed on a given area, we can follow these steps:
### Step 1: Calculate the number of moles of surfactant
Given:
- Molarity (C) = 1 mM = \(1 \times 10^{-3}\) M
- Volume (V) = 10 mL = \(10 \times 10^{-3}\) L = \(1 \times 10^{-2}\) L
Using the formula for moles:
\[
\text{Number of moles} = \text{Molarity} \times \text{Volume}
\]
\[
\text{Number of moles} = (1 \times 10^{-3} \, \text{mol/L}) \times (1 \times 10^{-2} \, \text{L}) = 1 \times 10^{-5} \, \text{mol}
\]
### Step 2: Calculate the number of surfactant molecules
Using Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{molecules/mol}\)):
\[
\text{Number of molecules} = \text{Number of moles} \times N_A
\]
\[
\text{Number of molecules} = (1 \times 10^{-5} \, \text{mol}) \times (6.022 \times 10^{23} \, \text{molecules/mol}) = 6.022 \times 10^{18} \, \text{molecules}
\]
### Step 3: Relate total area to the area occupied by one molecule
Given:
- Total area (A) = 0.24 cm²
The total area can be expressed as:
\[
A = \text{Area per molecule} \times \text{Number of molecules}
\]
Let \(a\) be the area occupied by one molecule:
\[
0.24 \, \text{cm}^2 = a \times (6.022 \times 10^{18})
\]
\[
a = \frac{0.24 \, \text{cm}^2}{6.022 \times 10^{18}} = 3.986 \times 10^{-20} \, \text{cm}^2
\]
### Step 4: Calculate the edge length of the cube
Since the surfactant molecules are assumed to be cube-shaped, the area occupied by one molecule can be expressed as:
\[
a = \text{edge length}^2
\]
Let \(l\) be the edge length:
\[
l^2 = 3.986 \times 10^{-20} \, \text{cm}^2
\]
\[
l = \sqrt{3.986 \times 10^{-20}} \approx 6.31 \times 10^{-10} \, \text{cm}
\]
### Step 5: Convert edge length to picometers
To convert centimeters to picometers:
\[
1 \, \text{cm} = 10^{10} \, \text{pm}
\]
\[
l \approx 6.31 \times 10^{-10} \, \text{cm} \times 10^{10} \, \text{pm/cm} = 63.1 \, \text{pm}
\]
### Final Answer
The edge length of the cube-shaped surfactant molecule is approximately **63.1 pm**.
