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