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

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

Serenity Allen · Accepted Answer

d)4.25g/cm3

Using the atomic masses of Cs and Br, determine the mass of one atom of CsBr. 133u plus 80u equals 213u. 213 g/mol convert pm to cm first. #1"pm" = 1xx^-10# #436.6"pm" *( 1xx-10cm)/(1"pm") = 4.366xx10^-8cm# Compute the unit cell's volume. #(4.366xx10^-8cm)^3 = 8.32245e-23cm^3# You now have to be aware of mass/volume equals density. Since we already know the mass, please convert amu to grams. 35369e-22 = 213amu #(3.5369e-22)/(8.32245e-23cm^3 ) = 4.24983027834"g/cm3 "# 3 sig figs = 4.25g/cm Option D is therefore accurate. An additional method is density = #(Z * mol.wt)/(a^3 * N_a)# where volume is equal to a^3 = atomic radius^3. #N_a# =Avogadro's number which is equal to #6.022*10^23# Z is the number of molecules per bbc. The bbc is usually 2 because bbc's are of atoms and elements have 1 atom which makes up a molecule. However, 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.

Naomi Aronson · Answer

undefined The density of CsBr is approximately 4.25g/cm^3 (Option d).

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