21)
In the above equation, if k is expressed in S m–1 and the concentration,
c in mol m–3 then the units of Lm are in S m2 mol–1 It may be noted that:
1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence
Lm(S cm2 mol–1) = 




1
3
1
 (S cm
)
1000 L m
 × molarity (mol L )
If we use S cm–1 as the units for k and mol cm–3, the units of
concentration, then the units for Lm are S cm2 mol–1 It can be calculated
by using the equation:
Lm (S cm2 mol–1) = 
1
3
 (S cm
) × 1000 (cm /L)
molarity (mol/L)


Both type of units are used in literature and are related to each
other by the equations:
1 S m2mol–1
= 104 S cm2mol–1   or
1 S cm2mol–1 = 10–4 S m2mol–1 Resistance of a conductivity cell filled with 0