Answer:
Alloy A is a BCC structure
Alloy B is SC structure
Alloy C is BCC structure
Explanation:
To solve for each of these three alloys we will need to do so by trial and error method to calculate the density and compare it to the value given in the problem. The density formula is;
ρ = nA/(V_c × N).
Where;
n is 1,2 and 4 for SC, BCC & FCC crystal structures respectively.
A is atomic weight
V_c is volume expressed as a³.
a is given by 2R,4R/√3, 2R√2 for SC, BCC & FCC crystal structures respectively with R being the atomic radius.
N is avogadros number of atoms and has a constant value of 6.022 × 10^(23) atoms/mol
A) Thus,
For metal alloy A, we are given;
A = 43.1 g/mol
If we assume it's a BCC structure, then;
n = 2
V_c = a³ = (4R/√3)³ = (4 × 1.22 × 10^(-8) × 1/√3)³ = 22.4756 × 10^(-24) cm³
N = 6.022 × 10^(23) atoms/mol
Plugging in the relevant values into the density formula, we have;
ρ = (2 × 43.1)/(22.4756 × 10^(-24) × 6.022 × 10^(23)) ≈ 6.4 g/cm³
This value is same as the one in the question, thus metal alloy A is a BCC crystal structure
B) For metal alloy B, we are given;
A = 184.4 g/mol
If we assume it's a SC structure, then;
n = 1
Thus;
V_c = a³ = (2 × 1.46 × 10^(-8))³ = 24.897 × 10^(-24) cm³
Plugging in the relevant values into the density formula, we have;
ρ = (1 × 184.4)/(24.897 × 10^(-24) × 6.022 × 10^(23)) ≈ 12.3 g/cm³
This value is same as the one in the question, thus metal alloy B is an SC crystal structure
C) For metal alloy C, we are given;
A = 91.6 g/mol
If we assume it's a BCC structure, then;
n = 2
V_c = a³ = (4R/√3)³ = (4 × 1.37 × 10^(-8) × 1/√3)³ = 31.67 × 10^(-24) cm³
Plugging in the relevant values into the density formula, we have;
ρ = (2 × 91.6)/(31.67 × 10^(-24) × 6.022 × 10^(23)) ≈ 9.6 g/cm³
This value is same as the one in the question, thus metal alloy C is an SC crystal structure
