An element has a cubic close packing (ccp) structure with an edge length of \(297\, \ pm\). Calculate the density of the element. (Given: Molar mass is \(107.9\, \ gmol^{-1}\))

\(2.7 × 10^4\, \ kgmol^{-1}\)

The correct answer is option 2. \(2.7 × 10^4\, \ kgmol^{-1}\)

We know the density, \(\rho \) can be calculated using the formula

\(\rho = \frac{Z × M}{N_A × a^3}\)

Given,

Edge length, \(a = 297\, \ pm = 297 × 10^{-12}\, \ m^3\)

Molar mass, \(M =  107.9\, \ gmol^{-1}\, \ = 107.9 × 10^{-3}\, \ kgmol^{-1}\)

Number of particles for cubic close packing (ccp) is, \(z = 4\)

Avogadro's number, \(N_A = 6.022 × 10^{23}\)

Applying these values in the above equation we get

\(\rho = \frac{4 × 107.9 × 10^{-3}\, \ kgmol^{-1}}{ 6.022 × 10^{23} × (297 × 10^{-12}\, \ m^3)^3}\)

or, \(\rho =\frac{0.4316}{6.022 × 26198073 × 10^{13}}\)

or, \(\rho =\frac{0.4316}{157764795.606 × 10^{13}}\)

or, \(\rho = 2.7357 × 10^4\)