Question

Fe - 10 atom % C austenite (fcc), having no Fe vacancies, has a lattice parameter of 4 Å. The density of austenite in g/cm³ is \(\underline{\hspace{2cm}}\) (round off to 2 decimal places).

Hint:

To calculate the density of a solid from its lattice parameter, use the formula \( \rho = \frac{Z \times A}{N_A \times a^3} \).

Answer 1

Correct Answer: 5.85

For austenite, the formula to calculate density is:
\[ \rho = \frac{Z \times A}{N_A \times a^3} \] Where: - \( Z = 4 \) (number of atoms per unit cell for fcc), - \( A = 55.8 \, \text{g/mol} \) (atomic weight of Fe), - \( N_A = 6.023 \times 10^{23} \, \text{atoms/mol} \), - \( a = 4 \times 10^{-10} \, \text{m} \) (lattice parameter). Substituting the values:
\[ \rho = \frac{4 \times 55.8}{6.023 \times 10^{23} \times (4 \times 10^{-10})^3} = 5.85 \, \text{g/cm}^3 \] Thus, the density of austenite is approximately \( 5.85 \, \text{g/cm}^3 \).