Suppose of sodium iodide is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to significant digits.

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

oh it is true if thats what you are asking

the answer is d. 74.02