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laila [671]
3 years ago
11

Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell

is 428.7 pm, what is the density of CsBr in g/cm3.
Chemistry
1 answer:
Oxana [17]3 years ago
5 0

Answer:

\mathbf {density \ d  =4.4845 \ g/cm^3}

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

\mathtt{ density \ d  = \dfrac{z \times molar\  mass  \ (M)}{ edge \ length \ (a)  \ \times avogadro's \ number \ (N)}}

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

\mathtt{ density \ d  = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}

\mathtt{ density \ d  = \dfrac{ 212.81}{47.4540533}}

\mathbf {density \ d  =4.4845 \ g/cm^3}

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