Answer :
To estimate the size of a molecule from the given data, we need to follow a series of steps. These include converting units, calculating the volume of the oil droplet, determining the area of the oil patch, and finally calculating the thickness of the oil film, which is an approximation of the diameter of the oil molecule. Here's the step-by-step solution:
1. Convert the mass to kilograms:
[tex]\[ \text{mass} = 0.9 \, \text{mg} \][/tex]
Since [tex]\( 1 \, \text{mg} = 1 \times 10^{-3} \, \text{g} = 1 \times 10^{-6} \, \text{kg} \)[/tex],
[tex]\[ \text{mass} = 0.9 \times 10^{-3} \, \text{g} = 0.9 \times 10^{-6} \, \text{kg} = 0.0009 \, \text{kg} \][/tex]
2. Convert the radius to meters:
[tex]\[ \text{radius} = 41.8 \, \text{cm} \][/tex]
Since [tex]\( 1 \, \text{cm} = 1 \times 10^{-2} \, \text{m} \)[/tex],
[tex]\[ \text{radius} = 41.8 \times 10^{-2} \, \text{m} = 0.418 \, \text{m} \][/tex]
3. Calculate the volume of the oil droplet using the density formula:
[tex]\[ \text{density} = \frac{\text{mass}}{\text{volume}} \][/tex]
Rearranging, we get:
[tex]\[ \text{volume} = \frac{\text{mass}}{\text{density}} \][/tex]
Given the density is [tex]\( 918 \, \text{kg/m}^3 \)[/tex],
[tex]\[ \text{volume} = \frac{0.0009 \, \text{kg}}{918 \, \text{kg/m}^3} \approx 9.80392156862745 \times 10^{-7} \, \text{m}^3 \][/tex]
4. Calculate the area of the oil patch:
The area of a circle is given by:
[tex]\[ \text{area} = \pi \times (\text{radius})^2 \][/tex]
Substituting the radius,
[tex]\[ \text{area} = \pi \times (0.418 \, \text{m})^2 \approx 0.548911634805823 \, \text{m}^2 \][/tex]
5. Calculate the thickness of the oil film, which approximates the diameter of an oil molecule:
[tex]\[ \text{thickness} = \frac{\text{volume}}{\text{area}} \][/tex]
Substituting the volume and area,
[tex]\[ \text{thickness} = \frac{9.80392156862745 \times 10^{-7} \, \text{m}^3}{0.548911634805823 \, \text{m}^2} \approx 1.7860655426069771 \times 10^{-6} \, \text{m} \][/tex]
Therefore, the diameter of the oil molecule is:
[tex]\[ \boxed{1.7 \times 10^{-6} \, \text{m}} \][/tex]
1. Convert the mass to kilograms:
[tex]\[ \text{mass} = 0.9 \, \text{mg} \][/tex]
Since [tex]\( 1 \, \text{mg} = 1 \times 10^{-3} \, \text{g} = 1 \times 10^{-6} \, \text{kg} \)[/tex],
[tex]\[ \text{mass} = 0.9 \times 10^{-3} \, \text{g} = 0.9 \times 10^{-6} \, \text{kg} = 0.0009 \, \text{kg} \][/tex]
2. Convert the radius to meters:
[tex]\[ \text{radius} = 41.8 \, \text{cm} \][/tex]
Since [tex]\( 1 \, \text{cm} = 1 \times 10^{-2} \, \text{m} \)[/tex],
[tex]\[ \text{radius} = 41.8 \times 10^{-2} \, \text{m} = 0.418 \, \text{m} \][/tex]
3. Calculate the volume of the oil droplet using the density formula:
[tex]\[ \text{density} = \frac{\text{mass}}{\text{volume}} \][/tex]
Rearranging, we get:
[tex]\[ \text{volume} = \frac{\text{mass}}{\text{density}} \][/tex]
Given the density is [tex]\( 918 \, \text{kg/m}^3 \)[/tex],
[tex]\[ \text{volume} = \frac{0.0009 \, \text{kg}}{918 \, \text{kg/m}^3} \approx 9.80392156862745 \times 10^{-7} \, \text{m}^3 \][/tex]
4. Calculate the area of the oil patch:
The area of a circle is given by:
[tex]\[ \text{area} = \pi \times (\text{radius})^2 \][/tex]
Substituting the radius,
[tex]\[ \text{area} = \pi \times (0.418 \, \text{m})^2 \approx 0.548911634805823 \, \text{m}^2 \][/tex]
5. Calculate the thickness of the oil film, which approximates the diameter of an oil molecule:
[tex]\[ \text{thickness} = \frac{\text{volume}}{\text{area}} \][/tex]
Substituting the volume and area,
[tex]\[ \text{thickness} = \frac{9.80392156862745 \times 10^{-7} \, \text{m}^3}{0.548911634805823 \, \text{m}^2} \approx 1.7860655426069771 \times 10^{-6} \, \text{m} \][/tex]
Therefore, the diameter of the oil molecule is:
[tex]\[ \boxed{1.7 \times 10^{-6} \, \text{m}} \][/tex]
