Question

If 0.015 M CaCl2 shows conductivity of 12,000 units, based on “total ion concentration,” what conductivity should be observed for 0.030 M Na2CO3? 12,000 32,000 24,000 36,000 30,000

Answer 1

If conductivity is purely based on total ion concentration this is straightforward.

CaCl₂ -> Ca²⁺+ 2Cl^- (three ions in solution, one of which can transfer 2 electrons, therefore with capacity to carry double the amount of charge)
For the other solution:
Na₂CO₃ -> 2Na⁺ + CO₃^2- (three ions in solution, one of which can transfer 2 electrons, therefore with capacity to carry double the amount of charge)
So both the solutions have the same conductivity for same concentrations.

But the concentration of the second solution is double than that of the first solution.

Therefore, conductivity=12000*2=24,000 units.