In water conservation, chemists spread a thin film of a certain inert material over the surface of water to cut down on the rate of evaporation of water in reservoirs, This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water of about 40 m2in area. Assuming that the oil forms a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in nanometers.

Extremely tiny clusters of atoms or molecules are often called nano-particles due to the small scale of their dimensions. The specific gravity of gold is 19.3 and the mass of one gold atom is 197 amu (where 1 amu = 1.66 x 10-24g). Calculate the number of gold atomsin a sphere-shaped nano-particle of gold that is 2.00 nanometers in diameter. [Hint: Recall that the volume of a sphere is (4/3) πR3where R is the sphere's radius and π = 3.1416.]

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Chemistry

Shiloh Sullivan

22 October, 11:31

In water conservation, chemists spread a thin film of a certain inert material over the surface of water to cut down on the rate of evaporation of water in reservoirs, This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water of about 40 m2in area. Assuming that the oil forms a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in nanometers.

Extremely tiny clusters of atoms or molecules are often called nano-particles due to the small scale of their dimensions. The specific gravity of gold is 19.3 and the mass of one gold atom is 197 amu (where 1 amu = 1.66 x 10-24g). Calculate the number of gold atomsin a sphere-shaped nano-particle of gold that is 2.00 nanometers in diameter. [Hint: Recall that the volume of a sphere is (4/3) πR3where R is the sphere's radius and π = 3.1416.]

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Riggs · Answer 1

length = 2.5 x 10⁻⁸ m

# atoms = 2500

Explanation:

The volume of oil spread is the area of the spread time its thckness

V = A x Thickness. This thickness is approximately the same as the the length of the molecule once we have a monolayer.

Performing our calculation and converting the mL to m³:

0.10 mL x 1 x 10⁻⁶ m³/mL = 1 x 10⁻⁷ m³

Thickness = V/A = 1 x 10⁻⁶ m³ / 40 m² = 2.5 x 10⁻⁸ m

For the second part assuming the specific gravity of the nanosphere is the same as its bulk density, we can establish an equation relating the specific gravity of a nanosphere to the value given of 19.3.

sp gravity = ρ gold / ρ H₂O ⇒ ρ gold = 19.3 x 1 g/cm³

So by definition of specific gravity, we have that

Number of atoms of gold x mass (amu) / vol nanosphere = 19.3

diameter = 2.0 nm x 1 m/10⁹ nm x 100 cm/m = 2.0 x 10⁻⁷ m

r = / 2 d = 1.0 x 10⁻⁷ m

Vnanosphere = 4/3 π r³ = 4/3 π (1 x 10⁻⁷ m) ³ = 4.2 x 10⁻²¹ m³

mass of gold atoms = number of gold stoms x mass (amu)

= # atoms x 197 x 1.66 x 10⁻²⁴ g/amu = # atoms x 3.3 x 10⁻²² g

sp gravity = ρ gold / ρ H₂O ⇒ ρ gold = 19.3 x 1 g/cm³

d = 19.3 = (# atoms x 3.3 x 10⁻²² g) / 4.2 x 10⁻²¹ cm³ / 1g/cm³

⇒ # atoms = 2500