Question

What is the molality of ions in a 0.253 M solution of Na₂SO₄, assuming the compound dissociates completely?

a) Calculations needed
b) Insufficient information
c) 0.253 m
d) 0.506 m

Answer 1

To calculate the molality of ions, we multiply the molarity of Na2SO4 by the number of ions produced per formula unit upon dissociation. However, molality requires the mass of the solvent, which is not provided, hence there is insufficient information to calculate ion molality.

Step-by-step explanation:

The question asks for the molality of ions in a 0.253 M solution of Na2SO4, assuming complete dissociation into its constituent ions. Since Na2SO4 completely dissociates into 2 Na+ ions and 1 SO42- ion, the molality of individual ions will be different from the molality of the compound itself.

To calculate the ion molality, we should consider that a 1 M solution of Na2SO4 would produce 3 moles of ions per liter (2 moles of Na+ and 1 mole of SO42-). Therefore, a 0.253 M solution produces 0.253 M * 3 = 0.759 moles of ions per liter of solution. However, molality is defined as moles of solute per kilogram of solvent. Without the mass of the solvent, we cannot convert molarity to molality. Therefore, the correct answer is (b) Insufficient information.