Asked by ChemGeek
the density of an unknown metal is 2.64 g/cm^3 and its atomic radius is 0.215 nm. It has a face-centered cubic lattice. What is the atomic weight.
Answer: 89.4 g/mol
how?
Answer: 89.4 g/mol
how?
Answers
Answered by
DrBob222
density = 2.64 g/cc
radius = 0.215 nm = 0.215E-9m = 0.215E-7 cm. It's far easier to change nm to cm at this point rather than finding volume in cubic nm and changing to cc then.
a = edge length = (4r/sqrt 2).
Then volume = a^3
mass of unit cell = volume x density = ??
Then substitute into the equation below and solve for atomic mass.
#atoms in unit cell x atomic mass = 6.022E23*mass unit cell
89.4 is the correct answer.
radius = 0.215 nm = 0.215E-9m = 0.215E-7 cm. It's far easier to change nm to cm at this point rather than finding volume in cubic nm and changing to cc then.
a = edge length = (4r/sqrt 2).
Then volume = a^3
mass of unit cell = volume x density = ??
Then substitute into the equation below and solve for atomic mass.
#atoms in unit cell x atomic mass = 6.022E23*mass unit cell
89.4 is the correct answer.
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