Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its th

Question

Eddi Din

4 days ago

15

Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its th

eoretical density?

Physics

2 answers:

Sav [3K] · Answer 1 · 2026-04-05T04:15:34+03:00

21.46 g/cm³

Further explanation

This calculation pertains to the topic of the arrangement of crystalline solids.

This question requires us to determine the theoretical density of platinum (Pt).

The formula for calculating theoretical density can be expressed as $\boxed{ \ \rho = \frac{(number \ of \ atoms/unit \ cell)(atomic \ mass)}{(volume \ of \ unit \ cell)(Avogadro's \ number)} \ }$

using the following symbols, i.e.,

$\boxed{ \ \rho = \frac{(n)(A)}{(V_C)(N_A)}\ }$

The given information includes:

an atomic radius of 0.1387 nm; therefore, $\boxed{R = 0.1387 \ nm \times \Big( \frac{10^{-9} \ m}{1 \ nm} \Big) \times \Big( \frac{10^2 \ cm}{1 \ m} \Big)}$

Consequently, $\boxed{ \ R = 0.1387 \times 10^{-7} \ cm \ = 1.387 \times 10^{-8} \ cm \ }$

the atomic mass is 195.08 g/mol

Since platinum exhibits an FCC crystal structure, n = 4 atoms, and $\boxed{ \ V_C = (2R \sqrt{2})^3 \ }$

Now, let’s calculate the unit cell volumes.

$\boxed{ \ V_C = (2(1.387 \times 10^{-8}) \sqrt{2})^3 \ }$

$\boxed{ \ V_C = (1.387 \times 10^{-8})^3 \times 16\sqrt{2} \ }$

$\boxed{ \ V_C = 6.038 \times 10^{-23} \ cm^3 \ }$

Insert all data into the general formula.

$\boxed{ \ \rho = \frac{(4 \ atoms/unit \ cell)(195.08 \ g/mol)}{(6.038 \times 10^{-23} \ cm^3/unit \ cell)(6.023 \times 10^{23} \ atoms/mol)} \ }$

Therefore, the theoretical density of platinum equates to 21.46 g/cm³ (rounded to two decimal places).

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Keywords: platinum, Pt, the FCC crystal structure, atomic radius, weight, mass, theoretical density, volume, face-centered cubic

inna [2.9K] · Answer 2 · 2026-04-05T04:17:48+03:00

The formula to apply is expressed as:

ρ = nA/VcNₐ
where
ρ signifies density
n represents the number of atoms within a unit cell (for FCC, n=4)
A indicates the atomic weight
Vc stands for the cubic cell volume equal to a³, with a being the side length (for FCC, a = 4r/√2, where r is the radius)\
Nₐ denotes Avogadro's number, which is 6.022×10²³ atoms/mol

Calculating the radius: r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
Then find a: a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
Then calculate V: V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

Now compute density:
ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]Finally, we get ρ = 21.46 g/cm³