Ratio of molar and equivalent conductance Al₂ (SO₄)₃ of a substance when 150g of it is solution :

To determine the ratio of molar conductance to equivalent conductance for aluminum sulfate, Al₂(SO₄)₃, we first need to understand a few key concepts related to conductance and how they apply to electrolytes in solution.

Understanding Molar and Equivalent Conductance

Molar conductance (Λm) refers to the conductance of a solution containing one mole of an electrolyte in a given volume, while equivalent conductance (Λe) is the conductance of a solution containing one equivalent of the electrolyte in the same volume. The relationship between these two can be expressed mathematically:

• Molar Conductance (Λm): Λm = κ × (V/n), where κ is the specific conductance, V is the volume of the solution, and n is the number of moles.
• Equivalent Conductance (Λe): Λe = κ × (V/e), where e is the number of equivalents.

Calculating Moles and Equivalents

For aluminum sulfate, Al₂(SO₄)₃, we need to find out how many moles and equivalents are present in 150 g of the substance. The molar mass of Al₂(SO₄)₃ can be calculated as follows:

• Aluminum (Al): 2 × 27 = 54 g/mol
• Sulfate (SO₄): 32 + (4 × 16) = 96 g/mol, so for three sulfates: 3 × 96 = 288 g/mol
• Total molar mass = 54 + 288 = 342 g/mol

Next, we calculate the number of moles in 150 g:

Number of moles (n) = mass (g) / molar mass (g/mol) = 150 g / 342 g/mol ≈ 0.438 moles.

Now, we need to determine the number of equivalents. Aluminum sulfate dissociates in solution as follows:

• Al₂(SO₄)₃ → 2 Al³⁺ + 3 SO₄²⁻

This means that one mole of Al₂(SO₄)₃ produces 2 equivalents of Al³⁺ and 3 equivalents of SO₄²⁻, totaling 5 equivalents per mole. Therefore, the number of equivalents (e) in 0.438 moles is:

Number of equivalents (e) = 0.438 moles × 5 equivalents/mole = 2.19 equivalents.

Finding the Ratio of Molar to Equivalent Conductance

Now that we have both the number of moles and equivalents, we can find the ratio of molar conductance to equivalent conductance:

Using the formulas:

• Λm = κ × (V/n)
• Λe = κ × (V/e)

Taking the ratio:

Ratio (Λm/Λe) = (κ × (V/n)) / (κ × (V/e)) = e/n.

Substituting the values we calculated:

Ratio = 2.19 equivalents / 0.438 moles ≈ 5.

Final Thoughts

The ratio of molar conductance to equivalent conductance for aluminum sulfate in this case is approximately 5. This reflects the fact that each mole of Al₂(SO₄)₃ produces multiple ions in solution, which contributes to its overall conductance. Understanding these relationships is crucial in fields like electrochemistry and analytical chemistry, where the behavior of electrolytes in solution is fundamental.