What is the molality of ions in a 0.573 m solution of (NH₄)₃PO₄ assuming the compound dissociates completely?

Question

What is the molality of ions in a 0.573 m solution of (NH₄)₃PO₄ assuming the compound dissociates completely?

Answer 1

Assuming the compound dissociates completely, you would get 4 particles (ions) from each particle (molecule) of (NH₄)₃PO₄ that was present. (NH₄)₃PO₄ ==> 3NH₄⁺ + PO₄^3- (4 particles).

If you have 0.573 m solution, that would be 0.573 moles of (NH₄)₃PO₄ per kg of solvent. The molality of ions would be 4 x 0.573 m = 2.29 m