[Borek]

OK, I got the question in PM. I did understood the question correctly:

Silver nitrate solution was added to the solution of monovalent metal iodide till the precipitation stopped. Mass of the produced silver iodide precipitate was identical to the mass of the iodide solution. What was the mass fraction of silver nitrate in its solution?

That's the question I tried to solve, and I don't see how it can be done. It is possible to calculate concentration of the iodide solution, although it requires assuming its density is 1 g/mL. This is unrealistic, as it yields iodide concentration in the range of 50%, sodium and potassium iodides of this concentration have density around 1.5 g/mL.

I wonder why you didn't want to post the original question on the forum - is it because you are not allowed to seek help?

[Rutherford]

Thanks anyway for the try. Didn't want to post because of private policy.

[AWK]

In a iodide aqua solution of a monovalent metal a solution of AgNO₃ is added until the separation of the AgI precipitate stopped. Mass of the obtained AgI is equal to the mass of the whole iodine solution. What was the mass share of AgNO₃ in its solution?
How to calculate something if there aren't any numbers given?
I started with the equation:
xMI+xAgNO₃ xAgI+xMNO₃
(108+127)x=m_r(of the iodide solution), but I don't have a clue what to do now.

This is a multisolution problem. Take for example 1 mole AgI as a resulting mass. From the stoichiometry of reaction a solution of iodide should contain 1 mole of (LiI, NaI, KI, RbI or ewentually NH₄I) and enough of water to the whole mass 108+127=235 g.

[Rutherford]

Then 1 mole of AgI is made from 170g of AgNO₃. M+127+m(H₂O)=235g, but I need the mass of water in the AgNO₃ solution.