A drop of oleic acid solution is placed on a tray with a light dust on the surface. The oleic acid spreads out one molecule thick and covers an area of 254 cm2. If you used 0.100% oleic acid and a dropper that took 20 drops to equal 1.0 mL, then what is the height of an oleic acid molecule in nm?

To find the height of an oleic acid molecule, we can treat the problem as one where the oleic acid forms a monolayer that spreads out over a certain area. We'll use the given information and solve this step by step.

### Given:
- The oleic acid solution is 0.100% (which means 0.100 g of oleic acid per 100 mL of solution).
- The total area covered by the monolayer is 254 cm².
- 20 drops are equivalent to 1.0 mL of solution.
- Each drop is a portion of the 0.100% solution.

### Step 1: Calculate the volume of one drop.
The dropper takes 20 drops to equal 1.0 mL, so the volume of one drop is:

\[
\text{Volume of one drop} = \frac{1.0 \, \text{mL}}{20} = 0.050 \, \text{mL}
\]

### Step 2: Calculate the mass of oleic acid in one drop.
Since the oleic acid concentration is 0.100% by mass, the mass of oleic acid in 1 mL of solution is:

\[
\text{Mass of oleic acid per mL} = 0.100 \, \text{g per 100 mL} = 0.001 \, \text{g per mL}
\]

Thus, the mass of oleic acid in one drop (0.050 mL) is:

\[
\text{Mass of oleic acid in one drop} = 0.001 \, \text{g/mL} \times 0.050 \, \text{mL} = 5 \times 10^{-5} \, \text{g}
\]

### Step 3: Estimate the volume of oleic acid in one drop.
The density of oleic acid is approximately \( 0.895 \, \text{g/mL} \). Using the mass of oleic acid in one drop, we can calculate its volume:

\[
\text{Volume of oleic acid in one drop} = \frac{\text{Mass}}{\text{Density}} = \frac{5 \times 10^{-5} \, \text{g}}{0.895 \, \text{g/mL}} = 5.59 \times 10^{-5} \, \text{mL}
\]

Since 1 mL = 1 cm³, the volume of oleic acid in one drop is:

\[
\text{Volume of oleic acid in one drop} = 5.59 \times 10^{-5} \, \text{cm}^3
\]

### Step 4: Calculate the height of the oleic acid molecule.
The oleic acid forms a monolayer covering an area of 254 cm². The thickness (height) of this monolayer is given by:

\[
\text{Height} = \frac{\text{Volume}}{\text{Area}} = \frac{5.59 \times 10^{-5} \, \text{cm}^3}{254 \, \text{cm}^2}
\]

\[
\text{Height} = 2.20 \times 10^{-7} \, \text{cm}
\]

### Step 5: Convert the height to nanometers.
Since \(1 \, \text{cm} = 10^7 \, \text{nm}\), the height in nanometers is:

\[
\text{Height} = 2.20 \times 10^{-7} \, \text{cm} \times 10^7 \, \text{nm/cm} = 2.20 \, \text{nm}
\]

Thus, the height of an oleic acid molecule is approximately **2.20 nm**.

---

Would you like more details or have any questions? Here are 5 related questions:

1. How does surface tension influence the spread of oleic acid on water?
2. What is the molecular structure of oleic acid and how does it relate to its height?
3. How would changing the concentration of oleic acid affect the thickness of the monolayer?
4. How does the density of oleic acid affect the outcome of this experiment?
5. Could this experiment be used to estimate the size of other molecules?

**Tip**: When working with nanometer-scale measurements, precision is crucial, so always account for all significant figures in calculations.

A drop of oleic acid solution is placed on a tray with a light dust on the surface. The oleic acid spreads out one molecule thick and covers an area of 254 cm². If you used 0.100% oleic acid and a dropper that took 20 drops to equal 1.0 mL, then what is the height of an oleic acid molecule in nm?

This problem involves calculating the height of an oleic acid molecule based on a spreading monolayer. Learn how to use geometric principles, volume, density, and unit conversions to solve for the molecular height in nanometers. Suitable for students in Grades 10-12.

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