# 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

Apr 9, 2017

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.

$1 \text{pm} = 1 {\times}^{-} 10$

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

Calculate the volume of unit cell

${\left(4.366 \times {10}^{-} 8 c m\right)}^{3} = 8.32245e-23 c {m}^{3}$

Now you must know

density = mass/volume

we have the mass already but convert amu to grams

213amu = 3.5369e-22

$\frac{3.5369e-22}{8.32245e-23 c {m}^{3}} = 4.24983027834 \text{g/cm3 }$

= 4.25g/cm3 (three sig figs)

Thus option d is correct

Another way is

density = $\frac{Z \cdot m o l . w t}{{a}^{3} \cdot {N}_{a}}$

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

${N}_{a}$ =Avogadro's number which is equal to $6.022 \cdot {10}^{23}$

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 .