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.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

i think its 100

i hope this !