A) 58
B) 57
C) 56
D) 55
Correct Answer: C
Solution :
[c] \[\frac{Fe}{O}=\frac{72.36}{\frac{56}{\frac{27.64}{16}}}=0.75=\frac{75}{100}=\frac{3}{4}\] \[\therefore F{{e}_{3}}{{O}_{4}}=3\times 56+4\times 16=32\] 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\[=\frac{1842}{232}\times 7=56\]
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