Question
How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$. Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$ Other given values are $$K_{sp_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{f_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$ Answer: $4.4\rm\,g$

My effort:
Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.

$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [Ag] = 1.8 \times 10 -9$$

Then I placed that value in formation eq.

$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$

$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$

$$x = 0.0288\,\rm M$$

and tried to find out $\ce{AgNO3}$ mass by using that by

$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$

That value doesn't seem to be the answer. What did I do wrong?

Question
How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$. Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$ Other given values are $$K_{\text{sp}_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{\mathrm{f}_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$ Answer: $4.4\rm\,g$

My effort:
Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.

$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [\ce{Ag}] = 1.8 \times 10 -9$$

Then I placed that value in formation eq.

$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$

$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$

$$x = 0.0288\,\rm M$$

and tried to find out $\ce{AgNO3}$ mass by using that by

$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$

That value doesn't seem to be the answer. What did I do wrong?

How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$.

Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$

Other given values are $$K_{sp_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{f_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$

Answer: $4.4\rm\,g$

My effort: Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.

$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [Ag] = 1.8 \times 10 -9$$

Then I placed that value in formation eq.

$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$

$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$

$$x = 0.0288\,\rm M$$

and tried to find out $\ce{AgNO3}$ mass by using that by

$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$

That value doesn't seem to be the answer. What did I do wrong?

Question
How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$. Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$ Other given values are $$K_{sp_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{f_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$ Answer: $4.4\rm\,g$

My effort:
Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.

$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [Ag] = 1.8 \times 10 -9$$

Then I placed that value in formation eq.

$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$

$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$

$$x = 0.0288\,\rm M$$

and tried to find out $\ce{AgNO3}$ mass by using that by

$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$

That value doesn't seem to be the answer. What did I do wrong?