Asked by Amanda L
Element X has a face-centered cubic unit cell structure. If the density of the element is measured to be 12.4 g/cm^3, what is the atomic radius of X? units are in pm (picometers)
MM: 102.9055 g/mol
I have done this 4 different times and came up with the same answer - 381pm, but it keeps saying it is wrong.
fcc = 4 atoms
102.9055 g/mol x 1 mol/6.022x10^23 atoms x 4 atoms/cell = 6.8353 X10^-22g
6.8353x10^-22g x cm^3/12.4g = 5.5123 cm^3
cubed root of 5.5123cm^3 = 3.806 x10^-8cm
3.806x10^-8cm x 10^-2m/cm x pm/10^-12m = 380.6pm
MM: 102.9055 g/mol
I have done this 4 different times and came up with the same answer - 381pm, but it keeps saying it is wrong.
fcc = 4 atoms
102.9055 g/mol x 1 mol/6.022x10^23 atoms x 4 atoms/cell = 6.8353 X10^-22g
6.8353x10^-22g x cm^3/12.4g = 5.5123 cm^3
cubed root of 5.5123cm^3 = 3.806 x10^-8cm
3.806x10^-8cm x 10^-2m/cm x pm/10^-12m = 380.6pm
Answers
Answered by
Scott
381 pm is the edge length of the cubic unit cell
the atomic radius is the edge length multiplied by ... √2 / 4
the atomic radius is the edge length multiplied by ... √2 / 4
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