Question

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 $\land_m$ for all concentrations and draw a plot between $\land_m$ and c ½. Find the value of $\land^0_m$.

Answer 1

Given,
$k$ = 1.237 × 10-2 S m-1 , c = 0.001 M
Then, $k$ = 1.237 × 10-4 S cm-1 , c ½ = 0.0316 M½

∴ $\land_m=\frac{\kappa}{c}$

= $\frac{1.237\times 10^{-4} S cm^{-1}}{0.001 mol L^{-1}} \times \frac{1000 cm^3}{L}$

=123.7 S cm2 mol-1

Given,
$k$ = 11.85 × 10-2 S m-1 , c = 0.010M
Then, $k$ = 11.85 × 10-4 S cm-1 , c ½ = 0.1 M½
$\therefore \land_m = \frac{\kappa}{c}$

$= \frac{11.85\times 10^{-4}S cm^{-1}}{0.010 mol L^{-1}}\times \frac{1000 cm^3}{L}$

$= 118.5 S cm^2mol^{-1}$
Given,
$k$ = 23.15 × 10-2 S m-1 , c = 0.020 M
Then, $k$ = 23.15 × 10-4 S cm-1 , c1/2 = 0.1414 M½
∴$\land_m = \frac{\kappa}{c}$

=$\frac{23.15\times 10^{-4}Scm^{-1}}{0.020 mol L^{1}}\times\frac{1000cm^3}{L}$

= 115.8 S cm2 mol-1
Given,
$k$ = 55.53 × 10-2 S m-1 , c = 0.050 M
Then, $k$ = 55.53 × 10-4 S cm-1 , c1/2 = 0.2236 M½
∴ $\land_m = \frac{\kappa}{c}$

= $\frac{55.53 \times 10^{-4}Scm^{-1}}{0.050 mol L^{-1}}\times\frac{1000 cm^3}{L}$

= 111.1 1 S cm2 mol-1

Given,
$k$ = 106.74 × 10-2 S m-1 , c = 0.100 M
Then, $k$ = 106.74 × 10-4 S cm-1 , c1/2 = 0.3162 M½
∴ $\land_m = \frac{\kappa}{c}$

= $\frac{106.74 \times 10^{-4}Scm^{-1}}{0.100 mol \; L^{-1}}\times\frac{1000cm^3}{L}$

= 106.74 S cm2 mol-1
Now, we have the following data:

c½/M½	0.0316	0.1	0.1414	0.2236	0.3162
$\land_m(S cm^2\, mol^{-1})$	123.7	118.5	115.8	111.1	106.74