Question

nacl has fcc structure calculated it's density if edges length is 562 pm and molar mass is 58.5 g/mal

Answer 1

As per the data given in the question,

Edge length of the unit cell (a) = 562 pm = 562×$10^{-10}$ cm

Molar mass of $NaCl$ (M) = 58.5 g/mol

Density of Cubic crystal (here $NaCl$) = (Z×M) ÷ ($a^{3}$×N$^{A}$)

where

Z = Number of formula units per unit cell

M = Formula mass of the compound

a = Edge length of the unit cell

N$_{A}$ = Avogadro's number

Unit cell of a face centred cubic system (fcc) contains 4 atoms. Thus for a fcc element, Z=4

So

Density = {(4×58.5) ÷ ([562×$10^{-10}$ $]^{3}$ × 6.02 × $10^{23}$ )}

⇒ Density = 2.189 g/$cm^{3}$

Hence density of $NaCl$ is 2.189 g/$cm^{3}$

Answer 2

Step 1: Given data

edge of the unit cell, $a= 562 pm=562\times 10^{-10} cm$

Molar mass of $NaCl, M=58.5g/mol$  

Density of Cubic crystal of $NaCl=\frac{(Z\times M)}{a^{3}\times N_{A} }$

where

$Z=$ Number of formula units per unit cell

$M=$ Formula mass of the compound

$a=$ Edge length of the unit cell

$N_{A}=$ Avogadro's number$=6.022\times10^{-23}$

Step 2: Calculating the density of $NaCl$

The unit cell of a face centred cubic system (fcc) contains 4 atoms.

Thus, for an fcc element, $Z=4$

Density of $Nacl=\frac{4\times 58.5}{(562\times 10^{-10})^{3}\times 6.022\times 10^{-23} }$

Density of  $NaCl= 2.189 g/cm^{3}$

Hence, density of $NaCl$ is$2.189g/cm^{3}$.

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