How many ml of 0.308-M Na3PO4 are required to deliver 46.1 mmol Na1+ ions?

1 mole of Na₃PO₄ contains 3 moles of Na⁺ ions.
No. of moles of Na⁺ ions = 46.1 mmol
No. of moles of Na₃PO₄ required = (46.1 mmol) × (1/3) = 46.1/3 mmol

Molarity of Na₃PO₄ = 0.308 M = 0.308 mmol/ml
Volume of Na₃PO₄ needed = (46.1/3 mmol) / (0.308 mmol/ml) = 49.9 ml