If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.

1. 10³ m²

2. 32.4 ml

3. 1.91 × 10²¹ molecules

Explanation:

Here is the complete question

1. Estimate the size of a one-molecule-thick oil film formed by spreading 1 ml of oil on the surface of the water. Assume that an oil molecule is roughly 10 nm in size. Show your work.

2. If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.

3. Estimate the number of molecules in 1 ml of oil assuming they’re 10 nm in size. What assumptions do you have to make? Show your work.

Solution

1. Since the film would cover an area, A, and would have a height which is the thickness of the molecule, h = 10 nm = 1 × 10⁻⁹ m, its volume is V = Ah. This volume also equals the volume of the oil. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, V = 1 × 10⁻⁶ m³.

The size of the oil drop is its area. So, A = V/h

= 1 × 10⁻⁶ m³ ÷ 1 × 10⁻⁹ m

= 10³ m²

2. The area of the tank is 10 in by 10 in = 100 in². Since we want to cover half the area, we require 100 in²/2 = 50 in² = 50 in² × (0.0254)² m²/in² = 0.0324 m².

If the thickness of oil is one molecule thick which is 10nm = 1 × 10⁻⁹ m, the volume of oil is then thickness × area = 1 × 10⁻⁹ m × 0.0324 m²

= 0.0324 × 10⁻⁹ m³

= 32.4 × 10⁻⁶ m³

= 32.4 ml since 1 × 10⁻⁶ m³ = 1 ml

3. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, we need to find the volume of one molecule. Since it is assumed to be a sphere, its volume is V' = πd³/6 where d = size of oil molecule = 10 nm = 1 × 10⁻⁹ m.

Let n be the number of molecules present in 1 ml, then nV' = 1 ml = 1 × 10⁻⁶ m³. So, n = 1 × 10⁻⁶ m³/V' = 1 × 10⁻⁶ m³ ÷ πd³/6 = 6 × 10⁻⁶ m³/πd³

Substituting d into the equation, we have

n = 6 × 10⁻⁶ m³/π(1 × 10⁻⁹ m)³

n = 6 × 10⁻⁶ m³/π × 10⁻²⁷ m³

n = 1.91 × 10²¹ molecules

resonant frequency

no

sodium is solid at rt