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Sagot :
Answer:
0.2762M Na+ in the solution
Explanation:
2.07g of sodium iodide Is Dissolved In 50.ML Of A 0.30M...
To solve this question we need to find the moles of sodium iodide, NaI, that are the same than the moles of sodium cation, Na+. The volume in liters of the solution is 0.050L. The molarity is:
Moles NaI = Moles Na+ -Molar mass NaI: 149.89g/mol-
2.07g NaI * (1mol / 149.89g) = 0.01381 moles NaI = Moles Na+
Molarity:
0.01381 moles Na+ / 0.0500L =
0.2762M Na+ in the solution
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