Question

CsBr crystallises in a body-centred cubic (bcc) lattice. The unit cell lengt is 436.6pm. Given that the atomic mass of Cs= 133u and that of Br=80u. What will be the density of CsBr? a) 0.425g/cm^3 b) 8.25g/cm^3 c) 42.5g/cm^3 d) 4.25g/cm^3

Answer 1

d)4.25g/cm3

Explanation:

Calculate the mass of one atom of CsBr from the atomic masses of Cs and Br

133u + 80u = 213u

213 g/mol

first convert pm to cm.

`1"pm" = 1xx^-10`

`436.6"pm" *( 1xx-10cm)/(1"pm") = 4.366xx10^-8cm`

Calculate the volume of unit cell

`(4.366xx10^-8cm)^3 = 8.32245e-23cm^3`

Now you must know

density = mass/volume

we have the mass already but convert amu to grams

213amu = 3.5369e-22

`(3.5369e-22)/(8.32245e-23cm^3 ) = 4.24983027834"g/cm3 "`

= 4.25g/cm3 (three sig figs)

Thus option d is correct

Another way is

density = `(Z * mol.wt)/(a^3 * N_a)`

Where a^3 = atomic radius^3 which is = volume

`N_a` =Avogadro's number which is equal to `6.022*10^23`

You would see that Z for bbc's are usually 2 but for this bbc Z = 1
because 1 bbc has 2atoms which here means 1atom of Cs and 1atom of Br giving a total of 1molecule and Z is the no. of molecule per bbc. The bbc is usually 2 because bbc's are of atoms and elements have 1 atom which make up a molecule .