Answer: option A, 0.065 M Na₃PO₄
Explanation:
1) First, notice that the solutes of every solution are ionic compounds with a high dissociation constant.
2) Then, you can assume full dissociation.
3) Also, lacking other information, and since conductity is the measure of the motion of charges (ions in the case of electrolytes), you can assume that for those solutions is valid the statement that "in many cases, conductivity is linked directly to the total dissolved solids".
4) Then, calculate the number of ions dissolved in each case to conclude which is the highest one and, with that, predict which electrolyte will conduct more electricity. Assume 1 liter for all of them, and use
M = n / V ⇒ n = M × V ⇒ n = M × 1 liter
5) A.0.065M Na₃PO₄.
Na₃PO₄ → 3Na⁺ + PO₄³⁻
1 mol → 4 moles
0.065 mol → 4×0.065 moles = 0.26 moles
6) B.0.075M (NH₄)₂CO₃.
(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻
1 mol → 3 moles
0.075 mol → 3×0.075 mol = 0.225 mol
7) C.0.10M LiBr.
LiBr → Li⁺ + Br⁻
1 mol → 2 mol
0.10 mol → 2×0.10mol = 0.20 mol
8) D.0.10M NaI.
NaI → Na⁺ + I⁻
1 mol → 2 mol
0.10 mol → 2×0.10 mol = 0.20 mol
9) E.0.10M KF.
KF → K⁺ + F⁻
1 mol → 2 mol
0.10 mol → 2×0.10 mol = 0.20 mol
10) Therefore the solution with more ions is the 0.065 M Na₃PO₄, with 0.26 moles, and that is the predicted solution with the highest electrical conductivity.
