Answer:
The density of the crystal is 10.22 g/cm³
Explanation:
Step 1: Data given
Molybdenum (Mo) has a BCC crystal structure
atomic radius = 0.1363 nm
atomic weight of 95.94 g/mol
Step 2: Calculate density of a bcc crystal
Density ρ= (nA *Mo)/(Vc*Na)
⇒For BCC, n = 2 atoms/unit cell, and realizing that V c = a ³
a = 4R/√3
Vc =(4R/√3)³
AMo = atomic weight = 95.94 g/mol
ρ = (n*A Mo)/((4R/√3)³ *Na)
⇒with n = 2 atoms/unit cell
⇒ with AMo = 95.94 g/mol
⇒ with Vc =(4R/√3)³ = (4*0.1363*10-7 cm)³ /(√3)³
⇒with Na = 6.022*10^23 atoms/mol
ρ =10.22 g/cm³
The density of the crystal is 10.22 g/cm³
Answer:
The density of the molybdenum BCC crystal structure is 10.22 g/cm^3
Explanation:
Density (D) of BCC crystal structure is given as:
D = (n × AW) ÷ (12.32r^3 × Na)
n is the number of atoms of molybdenum per unit cell = 2
AW is the atomic weight of molybdenum = 95.94 g/mol
r is the atomic radius of molybdenum = 0.1363 nm = 0.1363×10^-9 m = 0.1363×10^-9 × 100 = 1.363×10^-8 cm
Na is Avogadro's number = 6.022×10^23 atoms/mol
D = (2×95.94) ÷ [12.32 × (1.363×10^-8)^3 × 6.022×10^23] = 191.88 ÷ 18.78 = 10.22 g/cm^3
