Answer: The density of metal X is \(7.19g/cm^3\)
Explanation:
We are given:
Atomic radius of metal X = \(1.20\times 10^2pm=120pm\)
To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:
\(a=2\sqrt{2}R\)
Putting values in above equation, we get:
\(a=2\sqrt{2}\times 20=339.4pm\)
To calculate the density of metal, we use the equation:
\(\rho=\frac{Z\times M}{N_{A}\times a^{3}}\)
where,
\(\rho\) = density
Z = number of atom in unit cell = 4 (FCC)
M = atomic mass of metal = 42.3 g/mol
\(N_{A}\) = Avogadro's number = \(6.022\times 10^{23}\)
a = edge length of unit cell = \(339.4pm=339.4\times 10^{-10}cm\) (Conversion factor: \(1cm=10^{10}pm\) )
Putting values in above equation, we get:
\(\rho=\frac{4\times 42.3}{6.022\times 10^{23}\times (339.4\times 10^{-10})^3}\\\\\rho=7.19g/cm^3\)
Hence, the density of metal X is \(7.19g/cm^3\)
