Ccm bromide crystallizes in a body centred cubic lattice the unit cell length is 4 36.6 picometre given that the atomic mass of caesium equal to 133 and that of bromine equal to 80 amu and avogadro being 6.023 into 23 density of ccm bromide s
Answers
Answer:
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
=
1
×
−
10
436.6
pm
⋅
1
×
−
10
c
m
1
pm
=
4.366
×
10
−
8
c
m
Calculate the volume of unit cell
(
4.366
×
10
−
8
c
m
)
3
=
8.32245
e
−
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
3.5369
e
−
22
8.32245
e
−
23
c
m
3
=
4.24983027834
g/cm3
= 4.25g/cm3 (three sig figs)
Thus option d is correct
Another way is
density =
Z
⋅
m
o
l
.
w
t
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 .
Explanation:
see the below attachment for your answer
