Excess aqueous nai was added to 50 ml of
aqueous cuno3 solution, and 16.1 g of cui
precipitate was formed. what was the molar concentration of cuno3 in the original
solution?

The molarity of the solution is 1.69M.

The reaction between the two reactants are:

CuNO₃+ NaI = CuI +NaNO₃.

So, 1 mole of CuNO₃ or copper nitrate produces 1 mole of CuI or Copper iodide.

Molecular weight of copper iodide = = 190.4.

So, moles of copper iodide produced here is == 0.08.

So, moles of copper nitrate that was present in the solution = 0.08.

Amount of solution used = 50ml.

So, in 50ml or solution, moles of copper nitrate present = 0.08 moles.

So, in 1 liter of solution, moles of copper nitrate present = 1.69 moles.

So, the molarity of the solution is 1.69M.

i just got this problem. it's 90.3

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