Electrochemistry Part 1

Given
Conductivity of KCl solution kKCl=0.14Sm−1
RKCl=4.19ohm
RHCl=1.03ohm
To find conductivity of HCl solution kHCl
We know that,
k=1/R⋅G*
where k is the conductivity, R is the resistance and C* is the cell constant.
As k is constant. Therefore, k and C* constant
∴‌‌kKCl⋅RKCl=kHCl⋅RHCl
Therefore, 0.14×4.19=k×1.03
or, k of HCl solution =‌

0.14×4.19

1.03

=0.5695Sm−1=56.95×10−2Sm−1
=57×10−2Sm−1