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 M^½

∴ \(\land_m=\frac{\kappa}{c}\)

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

=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 M^½
\(\therefore \land_m = \frac{\kappa}{c}\)

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

\(= 118.5 S cm^2mol^{-1}\)
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 M^½
∴\(\land_m = \frac{\kappa}{c}\)

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

= 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 , c^1/2 = 0.2236 M^½
∴ \(\land_m = \frac{\kappa}{c}\)

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

= 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  , c^1/2 = 0.3162 M^½
∴ \(\land_m = \frac{\kappa}{c}\)

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

= 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