NCERT Solutions
Class 12 Chemistry
Electrochemistry

Q.10
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Concentration/M 0.001 0.010 0.020 0.050 0.100
102 × κ/S m–1 1.237 11.85 23.15 55.53 106.74
Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of Λ0m.
κ = 1.237 × 10−2 S m−1, c = 0.001 M
Then, κ = 1.237 × 10 −4 S cm−1, c½ = 0.0316 M1/2
Therefore, Λm = (κ/c)
= [(123.7 x 10-4 Scm-1)/ (0.001 molL-1)] x (1000cm3/L)
= [(123.7 Scm2mol−1
Given,
κ = 11.85 × 10−2 S m−1, c = 0.010M
Then, κ = 11.85 × 10 −4 S cm−1, c½ = 0.1 M1/2
Therefore, Λm = (κ/c)
= [(11.85 x 10-4 S cm-1)/ (0.010 mol L-1)] x (1000 cm3/L)
=118.5Scm2 mol-1 Given, κ = 55.53 × 10−2 S m−1, c = 0.050 M
Then, κ = 55.53 × 10 −4 S cm−1, c1/2 = 0.2236 M1/2
Therefore, κ = (κ/c)
= [(55.53 x 10-4 S cm-1)/ (0.050 mol L-1)] (1000 cm3/L)
= 111.11 S cm2 mol−1 Given, κ = 106.74 × 10−2 S m−1, c = 0.100 M
Then, κ = 106.74 × 10−4 S cm−1, c1/2 = 0.3162 M1/2
Therefore, Λm = (κ/c)
= [(106.74 x10-4) Scm-1/ (0.100) mol L-1]/ [1000 cm3/L)
= 106.74 S cm2 mol-1
Now the following data,
Since the line interrupts Λm at 124.0 S cm2mol-1, Λm0 =124.0 S cm2mol-1