02 mol L–1
= 1000 × 0 02 mol m–3 = 20 mol m–3
Molar conductivity
=

m
c

=
–3
–1
–3
248 × 10
 S m
20 mol m
 = 124 × 10–4 S m2mol–1
Alternatively,
k =
1 29 cm–1
520 
 = 0 248 × 10–2 S cm–1
Example 2