Final answer:
The degree of dissociation of Na₂SO₄, when comparing isotonic solutions with glucose, is calculated to be 75% using the van't Hoff factor. This takes into account the number of particles produced upon dissociation. However, there seems to be a discrepancy because the question mentions an apparent degree of dissociation of 85%.
Step-by-step explanation:
To determine the apparent degree of dissociation (α) of Na₂SO₄, we compare the isotonic solutions of Na₂SO₄ and glucose. Isotonic means that the solutions have the same osmotic pressure; therefore, they must have the same number of particles in solution. Since glucose does not dissociate, its 0.010M solution contains 0.010 mol of particles per liter.
Na₂SO₄ is a strong electrolyte and dissociates into 2 Na+ ions and 1 SO42- ion. If it were to fully dissociate, a 0.004M Na₂SO₄ solution would produce 0.004M x 3 = 0.012M of particles. However, the 0.004M Na₂SO₄ solution is isotonic with a 0.010M glucose solution, indicating incomplete dissociation.
To find the degree of dissociation (α), we use the formula:
α = (i - 1) / (n - 1), where i is the van't Hoff factor (i.e., the observed number of particles from one formula unit) and n is the number of particles expected upon complete dissociation. We know that i = total number of particles in isotonic solution / concentration of Na₂SO₄ = 0.010M / 0.004M = 2.5. Since n = 3 for Na₂SO₄ (2 Na+ and 1 SO42-),
α = (2.5 - 1) / (3 - 1) = 1.5 / 2 = 0.75 or 75%
However, the question states that the degree of dissociation is 85%. This could be a discrepancy that needs clarification, as the calculated degree does not match the stated answer. The calculation based on isotonic conditions suggests a 75% dissociation rate rather than 85%.