1. 1
+
-
The distance between the Li and
Cl ions in LiCl is 0.257 nm. Use
this and the molecular mass of LiCl
42.4 to compute the
density of LiCl.
g
mol
Solution
2. 2
0
3
0
23
0
We consider each ion to occupy
a cubic volume of side r . The
ions occupy a volume of 2 ,
where 6.02 10 is
Avogadro’s number. The density
is thus related to equilibrium
spacing r by
2
A
A
N r
N
M M
V N
3
0
(1)
Ar
23
7
0
Using
42.4 ,
6.022 10 / ,
0.257 10
in eq.(1), we get,
A
g
M
mole
N X mole
r X cm
3. 3
3
0
23
7 3
2
3
3
2
42.4
2 (6.022 10 / )
(0.257 10 )
207.39 10
2.0739
,
A
M M
V N r
g
mole
X X mole X
X cm
Now
g
X
cm
g
cm
