Step-by-step explanation:
We have four ionic compounds and with high dissociation constants, so we can assume that they are soluble and full dissociation.
We can also assume that the electrical conductivity is proportional to the number ions dissolved in the solution.
We have to find the number of ions dissolved for each solution supposing that we have 1 L of solution.
a) 0.10 M of LiBr:
LiBr ---> Li⁺ + Br⁻
1 mol of LiBr will dissolve into 2 mols of ions.
1 mol of LiBr = 2 moles of ions
Volume of solution = 1 L
molarity = 0.10 M
Molarity = moles of LiBr/(Volume of solution in L)
moles of LiBr = molarity * volume of solution in L
moles of LiBr = 0.10 M * 1 L
moles of LiBr = 0.10 moles
Since 1 mol of LiBr will be converted into 2 moles of ions.
moles of ions = 0.10 moles of LiBr * 2 moles of ions/(1 mol of LiBr)
moles of ions = 0.20 moles
b) 0.045 M of (NH₄)₂CO₃:
(NH₄)₂CO₃ ---> 2 NH₄⁺ + CO₃²⁻
1 mol of (NH₄)₂CO₃ = 3 moles of ions
molarity = 0.045 M
moles of (NH₄)₂CO₃ = 0.045 moles/L * 1 L
moles of (NH₄)₂CO₃ = 0.045 moles
moles of ions = 0.045 moles of (NH₄)₂CO₃* 3 moles of ions/(1 mol of (NH₄)₂CO₃)
moles of ions = 0.135 moles
c) 0.10 M of NaI:
NaI ---> Na⁺ + I⁻
1 mol of NaI = 2 moles of ions
molarity = 0.10 M
moles of NaI = molarity * volume of solution in L
moles of NaI = 0.10 M * 1 L
moles of NaI = 0.10 moles
moles of ions = 0.10 moles of NaI * 2 moles of ions/(1 mol of NaI)
moles of ions = 0.20 moles
d) 0.045 M of Al₂(SO₄)₃:
Al₂(SO₄)₃ ----> 2 Al³⁺ + 3 SO₄²⁻
1 mol of Al₂(SO₄)₃ = 5 moles of ions
molarity = 0.045 M
moles of Al₂(SO₄)₃ = 0.045 moles/L * 1 L
moles of Al₂(SO₄)₃ = 0.045 moles
moles of ions = 0.045 moles of Al₂(SO₄)₃* 5 moles of ions/(1 mol of Al₂(SO₄)₃)
moles of ions = 0.225 moles
The solutions with the greatest amount of ions is the Al₂(SO₄)₃ (0.225 moles) so we can assume that it will have the highest conductivity.
Answer: 0.045 M Al₂(SO₄)₃