The conductivity oh 0.1 N NaOH solution is 0.022 S cm−1. To this solution, an equal volume of 0.1 N HCl solution is added which results into a decrease of the conductivity of solution to 0.0055 S cm−1. The equivalent conductivity of NaCl solution is S cm2 equiv−1 is:
0.0055
0.11
110
55
A
0.11
B
0.0055
C
110
D
55
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Solution
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NaOH + HCl gives 2 free ions So, Λ=K×1000N
Where ,N= equivalent conductivity N= normality
Putting the values and evaluating, N=2×0.0055×10000.1=110Scm2eq−1
So, the equivalent conductivity of NaCl solution is 110Scm2eq−1
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