Given,
κ = 1.237 × 10−2 Sm−1, c = 0.001 M
Then, κ = 1.237× 10−4 S cm−1, c½ = 0.0316 M1/2
= 123.7 S cm2 mol−1
Given,
κ = 11.85 × 10−2 Sm−1, c = 0.010M
Then, κ = 11.85 × 10−4 Scm−1, c½ = 0.1 M1/2
= 118.5 S cm2 mol−1
Given,
κ = 23.15 × 10−2 Sm−1, c = 0.020 M
Then, κ =23.15 × 10−4 S cm−1, c1/2 = 0.1414 M1/2
= 115.8 S cm2 mol−1
Given,
κ = 55.53 × 10−2 Sm−1, c = 0.050 M
Then, κ = 55.53× 10−4 S cm−1, c1/2 = 0.2236 M1/2
= 111.1 1 S cm2 mol−1
Given,
κ = 106.74 × 10−2 Sm−1, c = 0.100 M
Then, κ = 106.74× 10−4 S cm−1, c1/2 = 0.3162 M1/2
= 106.74 S cm2 mol−1
Now, we have the following data:
Since the line interrupts
at 124.0 S cm2 mol−1, 
= 124.0 S cm2 mol−1.