[Rutherford]

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.

[Hunter2]

n₁=m₁/(x*M₁) = m₂/(y*M₂) = n₂

M is given by the silver salts and m₁ = m₂, x,y = 1 in this case

[Rutherford]

"Mass of the obtained AgI is equal to the mass of the whole iodine solution."
I don't understand your way.

[Borek]

n₁=m₁/(x*M₁) = m₂/(y*M₂) = n₂

M is given by the silver salts and m₁ = m₂, x,y = 1 in this case

It always helps to explain symbols used, otherwise we are forced to make guesses. I suppose you mean m₁ and m₂ are masses of AgNO₃ and AgI, if so, m₁≠m₂.

Perhaps I am missing something, but I don't see how to solve the problem. I can calculate concentration of iodide, but no idea how to calculate concentration of silver nitrate.

[whylolzz]

Hi, I tried attempting the question and reached an answer of some sort I think there is just enough material to work with to get an answer.
Do you have the actual answer you can provide so I can check whether it was right? Don't want to write it all out if it's wrong.

[Rutherford]

Sadly I don't have , but it would be nice if you write how did you solve this.

[whylolzz]

Actually, the question is quite vague. Is it stating that the mass of metal ['M' + AgNO₃] = [Mass of iodine solution]?

[Borek]

The way I read it it states "mass of the precipitated AgI = initial mass of the iodide solution".

[Rutherford]

Yes, it is so.

[whylolzz]

Yes, but when it says iodide solution, is the iodide solution just [MI] or [MI + AgNO₃].

[Borek]

MI IMHO.

[whylolzz]

$$ Let\;mass\;of\;AgNO_{3}=x
\\
\\
Let\;mass\;of\;MI=AgI=y
\\
\\
n_{AgNO_{3}}=\frac{x}{169.88}
\\
\\
AgNO_{3}\;\;is\;the\;limiting\;reagent\;as\;the\;question\;implies
\\
\\
n_{AgI}=\frac{y}{234.77}
\\
\\
\;\;But\;n_{AgI}=n_{AgNO_{3}}
\\
\\
\therefore\;\frac{x}{169.88}=\frac{y}{234.77}
\\
\\
\frac{x}{y}=0.72
\\
\\
\;hence\;72\%\;of\;the\;solution\;is\;AgNO_{3} $$

this is what I get as I interpret the question. However, I'm not sure whether the mass share is including AgNO₃

[Borek]

Looks to me like you are mistaking mass of the solution with mass of the solute. Yes, mass of the AgNO₃ is 72% of the mass of AgI that can be produced from that AgNO₃, but it has nothing to do with the question.

Note: don't use LaTeX for whole post, use it just for formulas (and not for chemical ones, unless really necessary). It looks better that way.

Radeford, can you post the original question without translating it to English?

[Rutherford]

How would you understand?

[Borek]

How would you understand?

Don't worry about that - if I will not understand it won't push you back, if I will understand, it may help. You have nothing to lose.