Question

Calculate the density of nacl if edge length of nacl unit cell is 564 pm

Answer 1

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

Answer 2

The density of NaCl crystal is $2.17g/cm^3$

Explanation:

To calculate the density of crystal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell = 4  (FCC)

M = atomic mass of NaCl = 58.5 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = $564pm=564\times 10^{-10}cm$    (Conversion factor:  $1cm=10^{10}pm$  )

Putting values in above equation, we get:

$\rho=\frac{4\times 58.5}{6.022\times 10^{23}\times (564\times 10^{-10})^3}\\\\\rho=2.17g/cm^3$

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