`10mL` of `1mM` surfactant solution forms a monolayer covering `0.24cm^(2)` on a polar substrate. If the polar head is approximated as a cube, what is its edge length ?

Options

A. `1.0p m`
B. `2.0p m`
C. `0.1p m`
D. 2.0 nm

Answer

To solve the problem step by step, we will follow the outlined procedure to find the edge length of the polar head of the surfactant.

### Step 1: Calculate the number of moles of surfactant in the solution
We have:
- Volume of surfactant solution = `10 mL`
- Concentration of surfactant = `1 mM` = `1 × 10^(-3) M`

Using the formula for moles:
\[ \text{Number of moles} = \text{Concentration (M)} \times \text{Volume (L)} \]
First, we convert the volume from mL to L:
\[ 10 \text{ mL} = 10 \times 10^{-3} \text{ L} = 0.01 \text{ L} \]
Now, calculate the number of moles:
\[ \text{Number of moles} = 1 \times 10^{-3} \text{ M} \times 0.01 \text{ L} = 1 \times 10^{-5} \text{ moles} \]

### Step 2: Convert moles to molecules
Using Avogadro's number, which is approximately \(6 \times 10^{23}\) molecules/mol:
\[ \text{Number of molecules} = 1 \times 10^{-5} \text{ moles} \times 6 \times 10^{23} \text{ molecules/mol} = 6 \times 10^{18} \text{ molecules} \]

### Step 3: Calculate the area occupied by one molecule
The total area covered by the surfactant is given as `0.24 cm²`. Therefore, the area occupied by one molecule is:
\[ \text{Area per molecule} = \frac{\text{Total area}}{\text{Number of molecules}} = \frac{0.24 \text{ cm}^2}{6 \times 10^{18}} = 4 \times 10^{-20} \text{ cm}^2 \]

### Step 4: Relate the area to the edge length of the cube
Since the polar head is approximated as a cube, the area \(A\) of one face of the cube is given by:
\[ A = a^2 \]
where \(a\) is the edge length of the cube. Thus, we have:
\[ a^2 = 4 \times 10^{-20} \text{ cm}^2 \]

### Step 5: Calculate the edge length \(a\)
Taking the square root of both sides:
\[ a = \sqrt{4 \times 10^{-20}} = 2 \times 10^{-10} \text{ cm} \]

### Step 6: Convert cm to picometers
Since \(1 \text{ cm} = 10^{10} \text{ pm}\):
\[ a = 2 \times 10^{-10} \text{ cm} \times 10^{10} \text{ pm/cm} = 2 \text{ pm} \]

### Final Answer
The edge length of the polar head is **2 picometers**.

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