The limiting molar conductivities of NaI, NaNO₃ and AgNO₃ are 12.7, 12.0 and 13.3 mS m² mol⁻¹, respectively (all at 25°C). The limiting molar conductivity of AgI at this temperature is ____________ mS m² mol⁻¹.

Solution

Chemistry ma’am · 4 steps

Build AgI from the salts you have: keep Ag⁺ and I⁻, and let Na⁺ and NO₃⁻ cancel out.

1. Kohlrausch's law. At infinite dilution every ion conducts on its own. So the limiting molar conductivity of a salt is the sum of the limiting conductivities of its cation and its anion.
2. Pick the combination. You need λ°_(Ag⁺) from AgNO₃ and λ°_(I⁻) from NaI. Adding these two also brings in Na⁺ and NO₃⁻, and subtracting NaNO₃ removes exactly that pair.

   Λ°_(AgI) = Λ°_(AgNO₃) + Λ°_(NaI) - Λ°_(NaNO₃)

3. Put in numbers. Λ°_(AgI) = 13.3 + 12.7 - 12.0 = 14.0 mS m² mol⁻¹.
4. So. The limiting molar conductivity of AgI is 14 mS m² mol⁻¹.

The common slip: 12.6. Subtracting NaI instead of NaNO₃ gives 13.3 + 12.0 - 12.7 = 12.6, a value that still contains NO₃⁻ and has lost I⁻.

The idea to keep. Write each salt as cation plus anion, then add and subtract salts until only the ions of your target salt are left. Check that every unwanted ion appears once with + and once with −.

Answer from the NTA final answer key. Solution by PrepWiser.