Calculate density of nacl its edge length of unit cell is 564pm and molar mass is 58.5
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given
a = 564pm = 564*10-10cm
fcc z= 4
molecular mass of nacl (M)=58.44g/mol
we know that
density D= MZ/NA* a3
D= 588.44 g * 4/ 6.022*1023 * (564 * 10-10)3
DENSITY (D)= 0.216367 g/cm3
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Answer: 2.16×106gm−32.16×106gm−3
The density is given by the formula zMa3NAzMa3NA
Given a=564ppma=564ppm.
Each unit cell of NaCl has 4 Na−− and 4 Cl−− ions →z=4→z=4
Avagadro's number NA=6.022×1023/molNA=6.022×1023/mol
The total mass of NaCl =22.99+34.34=58.5g/mol=22.99+34.34=58.5g/mol
⇒⇒ Density ρ=zMa3NAρ=zMa3NA=4×58.5(564×10−12)36.022×1023=4×58.5(564×10−12)36.022×1023gm−3≈2.16
The density is given by the formula zMa3NAzMa3NA
Given a=564ppma=564ppm.
Each unit cell of NaCl has 4 Na−− and 4 Cl−− ions →z=4→z=4
Avagadro's number NA=6.022×1023/molNA=6.022×1023/mol
The total mass of NaCl =22.99+34.34=58.5g/mol=22.99+34.34=58.5g/mol
⇒⇒ Density ρ=zMa3NAρ=zMa3NA=4×58.5(564×10−12)36.022×1023=4×58.5(564×10−12)36.022×1023gm−3≈2.16
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