When 0.1 mol (CoCl₃(NH₃)₅) is treated with excess of (AgNO₃), 0.2 mol of (AgCl) are obtained. The conductivity of the solution will correspond to:
a) (0.1, Ω⁻¹cm⁻¹)
b) (0.2, Ω⁻¹cm⁻¹)
c) (0.3, Ω⁻¹cm⁻¹)
d) (0.4, Ω⁻¹cm⁻¹)