how many unit cells are present in a) 2.5 gram cubic crystal of KCL b) with each edge of crystal?
I solved a), ans- 5.05×10²¹
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Hey dear,
● Answer -
n = 5.052×10^21 cells
● Explaination -
In cubic structure, one unit cell of KCl contains 4 KCl molecules.
Thus, mass of one unit cell is calculated by -
m = 4×74.5 / 6.022×10^23
No of unit cells in 2.5 g KCl,
n = W / m
n = 2.5×6.022×10^23 / 298
n = 5.052×10^21 cells
Therefore, 5.052×10^21 unit cells are present in in 2.5 g KCl.
Hop3 this help you...
● Answer -
n = 5.052×10^21 cells
● Explaination -
In cubic structure, one unit cell of KCl contains 4 KCl molecules.
Thus, mass of one unit cell is calculated by -
m = 4×74.5 / 6.022×10^23
No of unit cells in 2.5 g KCl,
n = W / m
n = 2.5×6.022×10^23 / 298
n = 5.052×10^21 cells
Therefore, 5.052×10^21 unit cells are present in in 2.5 g KCl.
Hop3 this help you...
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