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At a particular temperature. the ratio of equivalent conductance to specific conductance of a 0.01 N NaCl solution is
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Verified Answer
The correct answer is:
$10^{5} \mathrm{cm}^{3}$
As we know that,
Equivalent conductance, ( $\lambda$ )
$=\frac{\text { Specific conductance }(K) \times 1000}{\text { Concentration }}$
or. $\quad \frac{\lambda}{K}=\frac{1000}{\text { conc. }}$
Or. $\frac{\lambda}{K}=\frac{1000 \Omega^{-1} \mathrm{cm}^{2} \mathrm{eq}^{-1}}{0.01 \Omega^{-1} \mathrm{cm}^{-1}}$
$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$
(Given, conc. $=0.01 \mathrm{N}$ )
Or,$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$
Equivalent conductance, ( $\lambda$ )
$=\frac{\text { Specific conductance }(K) \times 1000}{\text { Concentration }}$
or. $\quad \frac{\lambda}{K}=\frac{1000}{\text { conc. }}$
Or. $\frac{\lambda}{K}=\frac{1000 \Omega^{-1} \mathrm{cm}^{2} \mathrm{eq}^{-1}}{0.01 \Omega^{-1} \mathrm{cm}^{-1}}$
$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$
(Given, conc. $=0.01 \mathrm{N}$ )
Or,$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$
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