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NCERT Class XII Chemistry
Chapter - The Solid State
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Question : 44 of 50
Marks: +1, -0
An element with molar mass 2.7×10−22.7 \times 10^{-2} kg mol−1^{-1} forms a cubic unit cell with edge length 405 pm. If its density is 2.7×1032.7 \times 10^{3} kg m−3^{-3}, what is the nature of the cubic unit cell?
Solution:  
Given M (molar mass of the element) =2.7×10−2= 2.7 \times 10^{-2} kg mol−1^{-1}
a (edge length) = 405 pm =405×10−12= 405 \times 10^{-12} m =4.05×10−10= 4.05 \times 10^{-10} m
d (density) =2.7×103= 2.7 \times 10^{3} kg m−3^{-3}
NAN_A (Avogadro’s number) =6.022×1023= 6.022 \times 10^{23} mol−1^{-1}
Using formula,
Density (d) =Z×Ma3×NA= \frac{Z \times M}{a^{3} \times N_{A}} or, Z=d×a3×NAMZ = \frac{d \times a^{3} \times N_{A}}{M}
or, Z =(2.7×103)(4.05×10−10)3(6.022×1023)2.7×10−2= \frac{ (2.7 \times 10^{3}) (4.05 \times 10^{-10})^{3} (6.022 \times 10^{23}) }{2.7 \times 10^{-2}} =4=4
Number of atoms of the element present per unit cell = 4. Hence, the cubic unit cell must be face-centred or cubic close packed (ccp).
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