Question
No. of atoms in Nichel unit cell = 4
Vslume of one unit all
[
=left(3 cdot 5 times 10^{-8}right)^{3} mathrm{cm}^{3}
]
Demidy of nickel all
( Rightarrow 4left(frac{5}{N}right) )
( frac{3 cdot 5 times 10^{-8}}{(3 cdot 5 times 1)^{3}}=9 )
( Rightarrow 4left(frac{5 q}{N}right)=9 timesleft(3 cdot 5 times 10^{-8}right)^{3} )
( begin{aligned} Rightarrow frac{236}{N} &=9 times 42 cdot 875 times 10^{-24} Rightarrow N &=frac{236}{982385.875 times 10^{-24}} &=6.11 times 10^{229} end{aligned} )
( = ) option ( (B) )

KABLA 24. By X-ray diffraction it is found that nickel (at. mass = 59 g mol-?), crystallizes with ccp. The edge length of the unit cell is 3.5 Å. If density of Ni crystal is 9.0 g/cmº. Then value of Avogadro's number from the data is: A. 6.05 x 1023 B. 6.11 x 1023 C. 6.02 x 1023 D. 6.023 x 1023 OTH bodol and tobedral
Solution
