Solid AB has NaCl type structure. If the radius of $A^{+}$ and $B^{-}$ are $1.5 \overset{˚}{A}$ and $2.5 \overset{˚}{A}$ respectively and formula mass of AB is $384 g /$ mole, what is the density of AB solid.
Take : Avogadro's number = 6 × 10²³atoms/mole

Question

Solid AB has NaCl type structure. If the radius of $A^{+}$ and $B^{-}$ are $1.5 \overset{˚}{A}$ and $2.5 \overset{˚}{A}$ respectively and formula mass of AB is $384 g /$ mole, what is the density of AB solid.

Take : Avogadro's number = 6 × 10²³atoms/mole

TestHub · Accepted Answer

Explain Question:

Calculating density of a NaCl-type solid from ionic radii.

Concept:

Density of Ionic Crystal (NaCl structure).

Solution:

In a rock-salt (NaCl) structure, the unit cell is Face-Centered Cubic (FCC) with $Z = 4$ .

The edge length ( $a$ ) is related to the radii of the cation ( $r^{+}$ ) and anion ( $r^{-}$ ) as they touch along the edge.

Calculation:

1. Edge length ( $a$ ): $a = 2 (r^{+} + r^{-}) = 2 (1.5 + 2.5) = 8.0 \overset{˚}{A} = 8 \times 1 0^{- 8} c m$ .

2. Formula:

$ρ = \frac{Z \times M}{N _{A} \times a ^{3}}$

where $ρ$ is density, $Z$ is the number of formula units per unit cell, $M$ is the molar mass, $N_{A}$ is Avogadro's number, and $a$ is the edge length.

3. Substitution:

$ρ = \frac{4 \times 384}{6 \times 1 0 ^{23} \times ( 8 \times 1 0 ^{- 8} ) ^{3}}$

$ρ = \frac{1536}{6 \times 1 0 ^{23} \times 512 \times 1 0 ^{- 24}}$

$ρ = \frac{1536}{307.2} = 5 g / c m^{3}$

Chemistry - Solid State Question with Solution | TestHub

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