Iron (Fe) has a density of 7.874 g/cm³ and crystallizes in a body-centered cubic structure. What is the atomic radius of iron? A)99 pm B)114 pm C)124 pm D)143 pm E)255 pm

Question

Quizplus · Accepted Answer

Using the formula for the density of a cubic crystal structure, we can solve for the atomic radius (r):

Density = (2 x atomic weight) / [(4/3) x pi x r^3]

Substituting the known values for iron, we get:

7.874 g/cm³ = (2 x 55.845 g/mol) / [(4/3) x pi x r^3]
r^3 = (2 x 55.845 g/mol) / [(4/3) x pi x 7.874 g/cm³]
r^3 = 0.1266 nm³
r = (0.1266 nm³)^(1/3)
r = 0.124 nm = 124 pm (answer choice C)

Iron (Fe) Has a Density of 7