Question

Metal M crystallizes into a fcc lattice with the edge length of \(4.0 × 10^{–8} cm\). The atomic mass of the metal is _______ g/mol. (Nearest integer) 

(Use :\( N_A=6.02×10^{23} mol^{-1}\),density of metal , \(M = 9.03 g cm^{- 3}\))

Answer 1

Given : fcc lattice ⇒ number of atoms per unit cell \(Z = 4\)

Edge length \(a = 4.0 \times 10^{-8}\, cm\)

Density \(d = 9.03\, g\,cm^{-3}\)

Avogadro number \(N_A = 6.02 \times 10^{23}\, mol^{-1}\)

\(\begin{array}{l} \text{Density formula : } d = \frac{Z \cdot M}{a^3 \cdot N_A} \end{array}\)

\(\begin{array}{l} M = \frac{d \cdot a^3 \cdot N_A}{Z} \end{array}\)

\(\begin{array}{l} M = \frac{9.03 \times (4.0 \times 10^{-8})^3 \times (6.02 \times 10^{23})}{4} \end{array}\)

\(\begin{array}{l} (4.0 \times 10^{-8})^3 = 64 \times 10^{-24} \end{array}\)

\(\begin{array}{l} M = \frac{9.03 \times 64 \times 10^{-24} \times 6.02 \times 10^{23}}{4} \end{array}\)

\(\begin{array}{l} M = \frac{9.03 \times 64 \times 6.02 \times 10^{-1}}{4} \end{array}\)

\(\begin{array}{l} M = \frac{347.8}{4} \approx 87 \end{array}\)

The atomic mass of the metal = 87 g/mol