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 `bidwedge__(eq)(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
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The correct Answer is:
B
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