How many ml of 0.308-M Na3PO4 are required to deliver 46.1 mmol Na1+ ions?
回答 (1)
2017-01-25 12:37 am
✔ 最佳答案
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
收錄日期: 2021-04-18 16:05:28
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