Solid State Question 262
Question: A mineral of iron contains an oxide containing 72.36% iron by mass and has a density of 5.2 g/cc. Its unit cell is cubic with edge length of 839 pm. What is the total number of atoms (ions) present in each unit cell- (Fe-56, O-16)
Options:
58
57
56
55
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{Fe}{O}=\frac{72.36}{\frac{56}{\frac{27.64}{16}}}=0.75=\frac{3}{4} $ $ \therefore Fe_3O_4=3\times 56+4\times 16=232 $ Density= $ \frac{Mass}{Volume} $ $ \Rightarrow $ Mass of 1 unit cell=Density $ \times N _{A}\times a^{3} $ amu $ =5.2\times 6.023\times 10^{23}\times {{(8.39\times {{10}^{-8}})}^{3}} $
$ =1842 $ amu No. of atoms $ =\frac{1842}{232}\times 7=56 $
BETA
