The conductivity of sodium chloride at 298 K ...

NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

Question : 10 of 32

Marks: +1, -0

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
$\frac{10^{2} \times \kappa}{\mathrm{S}\,\mathrm{m}^{-1}}$	1.237	11.85	23.15	55.53	106.74

Calculate

\Lambda_m

for all concentrations and draw a plot between

\Lambda_m

and

c^{1/2}

.Find the value of

\Lambda^{0}_m

Solution:

1\,\mathrm{S}\,\mathrm{cm}^{-1}=100\,\mathrm{S}\,\mathrm{m}^{-1}

\frac{1\,\mathrm{S}\,\mathrm{cm}^{-1}}{100\,\mathrm{S}\,\mathrm{m}^{-1}}=1

(unit conversion factor)

Conc. (M)	$\kappa(S\,\mathrm{m}^{-1})$	$\kappa(S\,\mathrm{cm}^{-1})$	$\Lambda_m = \frac{1000 \times \kappa}{\text{Molarity}} (S\,\mathrm{cm}^2$ $\mathrm{mol}^{-1} )$	$c^{1/2} (\mathrm{M}^{1/2} )$
$10^{-3}$	$1.237 \times$ $10^{-2}$	$1.237 \times$ $10^{-4}$	$\frac{1000 \times 1.237 \times 10^{-4}}{10^{-3}}=123.7$	$0.0316$
$10^{-2}$	$11.85 \times$ $10^{-2}$	$11.85 \times$ $10^{-4}$	$\frac{1000 \times 11.85 \times 10^{-4}}{10^{-2}}=118.5$	$0.100$
$2\times10^{-2}$	$23.15 \times$ $10^{-2}$	$23.15 \times$ $10^{-4}$	$\frac{1000 \times 23.15 \times 10^{-4}}{2 \times 10^{-2}}=115.8$	$0.141$
$5\times10^{-2}$	$55.53 \times$ $10^{-2}$	$55.53 \times$ $10^{-4}$	$\frac{1000 \times 55.53 \times 10^{-4}}{5 \times 10^{-2}}=111.1$	$0.224$
$10^{-1}$	$106.74 \times$ $10^{-2}$	$106.74 \times$ $10^{-4}$	$\frac{1000 \times 106.74 \times 10^{-4}}{10^{-1}}=106.7$	$0.316$

\Lambda^{0}

= Intercept on the

\Lambda_m

axis

= 124.0 \,\mathrm{S}\,\mathrm{cm}^{2}\,\mathrm{mol}^{-1}

(on extrapolation to zeroconcentration).

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