Question

The conductivity of `"0.1N NaOH"` solution is `0.022 S cm^(-1)`. To this solution equal volume of `" 0.1 N HCl` solution is added which results into decrease of conductivity of solution to `0.0055 S cm^(-1)`. The equivalent conductivity of `NaCl` solution in `S cm^(2)" equiv"^(-1)` is `:`
A. `0.011`
B. `110`
C. `0.0055`
D. `55.0`

Answer 1

Correct Answer - 2
Normality of resulting solution `=(0.1V)/(2V)=0.05N`
`wedge_(eq)=(Kxx1000)/(N)=(0.0055xx1000)/(0.05)=110`