the unit cell length of NaCl crystal is 564 pm. its density whould be
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Answer: 2.16×106gm−3
The density is given by the formula zMa3NA
Given a=564ppm.
Each unit cell of NaCl has 4 Na− and 4 Cl− ions →z=4
Avagadro's number NA=6.022×1023/mol
The total mass of NaCl =22.99+34.34=58.5g/mol
⇒ Density ρ=zMa3NA=4×58.5(564×10−12)36.022×1023gm−3≈2.16×106gm−3
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