A solution is prepared by dissolving 20.0 g of NaI in enough water to make 300.0 mL of solution. How many moles of ions are in 25.0 mL of this solution?

Question

Quizplus · Accepted Answer

First, calculate the moles of NaI in the entire solution: Molar mass of NaI = 149.89 g/mol. Moles of NaI = 20.0 g / 149.89 g/mol = 0.1335 mol. Since NaI dissociates into Na+ and I- ions, the total moles of ions in 300 mL is 0.1335 mol * 2 = 0.267 mol. For 25 mL, the moles of ions = (25 mL / 300 mL) * 0.267 mol = 0.0222 mol.

A Solution Is Prepared by Dissolving 20