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 =