A gold atom is a sphere with a diameter of 272 pm and a mass of 3.27 x 10-13 ng. Calculate the density of one gold atom (in g/cm3). Note the Vsphere = 4/3 π r3.

Question

Zaria Dalton

Chemistry

17 September, 17:22

A gold atom is a sphere with a diameter of 272 pm and a mass of 3.27 x 10-13 ng. Calculate the density of one gold atom (in g/cm3). Note the Vsphere = 4/3 π r3.

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Ahmed Ferguson · Answer 1

Density = mass/volume

Mass is 3.27 x 10^-13ng.

Since diameter is 272 pm, the radius is half of that: 137pm.

Now solve the volume: Vsphere=4/3π (137^3) ≈1091317pm³

Then solve density:

3.27x10^-13ng/1091317pm³ ≈3.00x10^-19 ng/pm³

Convert units: 3.00x10^-19ng x (1 gram/1x10^9 grams) x (1 pm/1x10^-10cm) ³=300g/cm³