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 \(107.9\, \ gmol^{-1}\))
Answer & explanation
Correct answer: option 2
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, \(\rho = 2.7357 × 10^4\)