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Question: Answered & Verified by Expert
At a particular temperature. the ratio of equivalent conductance to specific conductance of a 0.01 N NaCl solution is
ChemistryElectrochemistryWBJEEWBJEE 2015
Options:
  • A $10^{5} \mathrm{cm}^{3}$
  • B $10^{3} \mathrm{cm}^{3}$
  • C $10 \mathrm{cm}^{3}$
  • D $10^{5} \mathrm{cm}^{2}$
Solution:
1198 Upvotes 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,$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$

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