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Let us discuss the crystal structure of $\text{NaCl}$ in detail.

In sodium chloride or $\text{NaCl}$, each ion is surrounded by six ions of the opposite charge. The neighbouring ions are located at the vertices of a regular octahedron. The larger chloride ions are arranged in a cubic array whereas sodium ions fill octahedral voids between them. Let us now solve the numerical using the formula; $\text{d=}\dfrac{\text{Z}\times \text{M}}{{{\text{a}}^{3}}\times {{\text{N}}_{\text{A}}}}$; where$\text{d}$ is density of the crystal, $\text{M}$ is the molar mass of solid, $\text{a}$ is the edge length or side of the cubic crystal, $\text{Z}$ is the effective number of atoms present in a unit cell or how many formula units are there in a unit cell and ${{\text{N}}_{\text{A}}}$ is Avogadro’s Number with a fixed value of $6.022\times {{10}^{23}}$.

Step (1) - Write the values given in the question first, $\text{Z=4}$(effective number of atoms present in a unit cell), $\text{a = 0}\text{.564 nm or a= 5}\text{.67}\times \text{1}{{\text{0}}^{-10}}\text{m}$ but we need to find$\text{d}$in$\text{g/c}{{\text{m}}^{3}}$, for that we need to convert metres to centimetres. We know that $1\text{ metre= 100 centimetres}$; so edge length will be equal to $\text{a=5}\text{.67}\times \text{1}{{\text{0}}^{-10}}\times \text{100 cm}$or $\text{a=5}\text{.67}\times \text{1}{{\text{0}}^{-8}}\text{ cm}$ and the molar mass of $\text{NaCl}$ is 58.5 grams (found above).

Step (2) – Apply the formula by putting all the values:

The formula will be now $\text{d=}\dfrac{4\times 58.5}{{{\left( 5.67\times {{10}^{-8}} \right)}^{3}}\times 6.022\times {{10}^{23}}}$

Step (3) – By calculations, find the value of $\text{d}$; $\text{d}$ is equal to 2.13$\text{g/c}{{\text{m}}^{3}}$.

The density of sodium chloride is 2.13 $\text{g/c}{{\text{m}}^{3}}$.