Solid State Question 274
Question: How many unit cells are present in a cube-shaped ideal crystal of $ NaCl $ of mass $ 1.00g $ -
[Atomic masses: $ Na=23,Cl=35.5 $ ]
Options:
A) $ 5.14\times 10^{21} $ unit cells
B) $ 1.28\times 10^{21} $ unit cells
C) $ 1.71\times 10^{21} $ unit cells
D) $ 2.57\times 10^{21} $ unit cells
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Since in $ NaCl $ type of structure 4 formula units form a cell. 58.5 gm. Of $ NaCl=6.023\times 10^{23} $ atoms 1 gm of atoms $ NaCl=\frac{6.023\times 10^{23}}{58.5} $ 4 atoms constitute 1 unit cell
$ \therefore =\frac{6.023\times 10^{23}}{58.5} $ atoms constitute $ =\frac{6.023\times 10^{23}}{58.5\times 4} $
$ =2.57\times 10^{21} $ unit cells.
BETA
