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:
A) 58
B) 57
C) 56
D) 55
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \frac{Fe}{O}=\frac{72.36}{\frac{56}{\frac{27.64}{16}}}=0.75=\frac{75}{100}=\frac{3}{4} $
$ \therefore Fe_3O_4=3\times 56+4\times 16=32 $ 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 {{(839\times {{10}^{-10}})}^{3}} $
$ =1842 $ amu No. of atoms $ =\frac{1842}{232}\times 7=56 $
BETA
