Paul 10.2

1. 1
0
+ --
0
3
Calculate the distance r between
the K and Cl ions in KCl, assuming
that each ion occupies a cubic volume
of side r . The molar mass of KCl is
74.55 and its density is 1.984 .
g g
mol cm
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
3
0 (1)
2
A
A
r
M
r
N 
    
3
23
Using
=1.984 ,
74.55 ,
6.022 10 /
in eq.(1), we get,
A
g
cm
g
M
mole
N X mole




3. 3
3
0
23
3
23 3
29 3
30 3
10
0
9
2
74.55
2 (6.022 10 / )
(1.984 )
3.12 10
3.12 10
31.2 10
31.2 10
3.12 10 3.12
A
M
r
N
g
mole
X X mole X
g
cm
X Xcm
X m
X m
r X m
X m nm











 
 