To solve this experiment, let's go through the questions step-by-step.
1) Calculate the volume of one oil drop.
You have a total volume of oil drops, which is 1.54 cm³, for 100 drops.
The volume of one oil drop \( V_{drop} \) can be calculated as: \[ V_{drop} = \frac{\text{Total volume of oil}}{\text{Number of drops}} = \frac{1.54 , \text{cm}^3}{100} = 0.0154 , \text{cm}^3. \]
2) Calculate the area of the oil patch.
The area \( A \) of a circular patch can be calculated using the formula: \[ A = \pi r^2, \] where \( r \) is the radius of the patch. The diameter of the patch is given as 17.5 cm, so the radius \( r \) is: \[ r = \frac{d}{2} = \frac{17.5 , \text{cm}}{2} = 8.75 , \text{cm}. \]
Now, calculating the area: \[ A = \pi (8.75 , \text{cm})^2. \] Calculating further: \[ A \approx 3.14159 \times 76.5625 , \text{cm}^2 \approx 240.35 , \text{cm}^2. \]
3) Calculate the thickness of the oil patch.
The thickness \( t \) of the oil patch can be calculated as follows: \[ t = \frac{V_{drop}}{A}, \] where \( V_{drop} \) is the volume of one drop, and \( A \) is the area of the patch.
Substituting the values we have: \[ t = \frac{0.0154 , \text{cm}^3}{240.35 , \text{cm}^2} \approx 0.0000641 , \text{cm} \approx 0.000641 , \text{mm} \text{ (or 0.641 μm)}. \]
4) Calculate the size of the oil molecule.
To estimate the size of one oil molecule, we can assume that the thickness of the oil patch is due to a monolayer of oil molecules. The volume of one oil molecule can be approximated as the thickness of the oil patch multiplied by the area of one drop:
Assuming the oil patch consists of a monolayer of oil molecules, we can express the volume of one molecule using: \[ V_{molecule} = \frac{V_{drop}}{N}, \] where \( N \) is the number of molecules in a drop.
Assuming an average size for an oil molecule (for example, around 1.5 nm = \( 1.5 \times 10^{-7} \) cm in diameter), the number of oil molecules in one drop can be approximated as:
Using the volume of one oil molecule based on the thickness: The volume of one molecule, if we assume it's roughly spherical and of diameter \( d \) is: \[ V_{molecule} = \frac{\pi (d/2)^3}{6}. \] Let’s estimate \( d \): Using the relation \( t \) and \( V_{molecule} \). In practical scenarios, you can often use Avogadro's number to find the effective number \( N \).
However, more concretely defining \( d \) requires additional approximations or data on specific molecules or an assumption about the arrangement of molecules.
The thickness (or height) of the oil film can directly be considered in estimating the length of molecules.
To calculate the approximate number of moles per drop (you could convert each drop to number of molecules existing in a volume following Avogadro's principle).
1 drop = \( 0.0154 , \text{cm}^3 \) Since you need detailed molecular size data, it typically ranges based on oil sources.
Thus, without specific molecular volume, you can conclude or factor with the rest of backend measurements of \( 3410^{23} \) and practical approximated sizes commonly stipulated.
The molecule calculations become much clearer with standard molecule properties and constants taken from specific material data.
Summary
- Volume of one oil drop = 0.0154 cm³
- Area of the oil patch = 240.35 cm²
- Thickness of the oil patch = 0.0000641 cm (or 0.641 μm)
- Size of the oil molecule would require further structural specific data estimates or properties corresponding to the oil.