The molar conductivities of infinite dilution for sodium iodide, sodium acetate and aluminium acetate are 12.69, 9.10 and 24.52 S cm2 mol−1 respectively at 25 °C. What is the molar conductivity of AlI3 at infinite dilution?

To find the molar conductivity of aluminum iodide (AlI3) at infinite dilution, we can use the concept of molar conductivities of its constituent ions. Molar conductivity at infinite dilution is the sum of the molar conductivities of the individual ions that make up the compound. In the case of AlI3, it dissociates into one aluminum ion (Al³⁺) and three iodide ions (I⁻) when dissolved in water.

Understanding the Components

First, let’s identify the ions produced by AlI3:

• 1 Al³⁺ ion
• 3 I⁻ ions

Known Molar Conductivities

From the data provided, we have the following molar conductivities at infinite dilution:

• Sodium iodide (NaI): 12.69 S cm² mol⁻¹
• Sodium acetate (NaC2H3O2): 9.10 S cm² mol⁻¹
• Aluminum acetate (Al(C2H3O2)3): 24.52 S cm² mol⁻¹

However, we need the molar conductivity of the ions relevant to AlI3. The molar conductivity of the iodide ion (I⁻) can be derived from sodium iodide, as it contains the same iodide ion. Sodium ions (Na⁺) are not needed for our calculation, but we can use the molar conductivity of NaI to find that of I⁻.

Calculating Molar Conductivity of Iodide Ion

The molar conductivity of NaI is the sum of the conductivities of Na⁺ and I⁻:

Let’s denote:

• Λ(Na⁺) = Molar conductivity of sodium ion
• Λ(I⁻) = Molar conductivity of iodide ion

Thus, we can express the molar conductivity of NaI as:

Λ(NaI) = Λ(Na⁺) + Λ(I⁻)

Since we don’t have the value for Λ(Na⁺), we can assume it is negligible compared to Λ(I⁻) for our purpose, or we can look for typical values. However, for simplicity, we can focus on the fact that we need the molar conductivity of I⁻, which we can estimate as being close to the value of NaI.

Finding Molar Conductivity of AlI3

Now, we can calculate the molar conductivity of AlI3 using the molar conductivities of its ions:

Λ(AlI3) = Λ(Al³⁺) + 3 × Λ(I⁻)

From the data, we know that the molar conductivity of Al³⁺ can be approximated from aluminum acetate, but it’s more common to find it from literature values. A typical value for Λ(Al³⁺) is around 120 S cm² mol⁻¹. Now, we can plug in the values:

Assuming Λ(I⁻) is approximately equal to Λ(NaI) = 12.69 S cm² mol⁻¹, we can calculate:

• Λ(Al³⁺) = 120 S cm² mol⁻¹
• Λ(I⁻) = 12.69 S cm² mol⁻¹

Now substituting these values into our equation:

Λ(AlI3) = 120 + 3 × 12.69

Λ(AlI3) = 120 + 38.07

Λ(AlI3) = 158.07 S cm² mol⁻¹

Final Result

Therefore, the molar conductivity of aluminum iodide (AlI3) at infinite dilution is approximately 158.07 S cm² mol⁻¹.