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In above question after formation of NaC...

In above question after formation of NaCl, further 0.1 N HCl is added, the volume of which is double to that of the first portion added, the conductivity increases to 0.018 `Scm^(-1)`. The value of `wedge__(aq)`(HCl) is [assume no change in equivalent conductivity of NaCl(aq)]:

A

330 S`cm^2eq^(-1)`

B

305 S`cm^2eq^(-1)`

C

415 S`cm^2eq^(-1)`

D

360 S`cm^2eq^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
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