21.6 g of silver coin is dissolved in ${\mathrm{HNO}}_3$. When $\mathrm{NaCl}$ is added to this solution, all silver is precipitated as $\mathrm{AgCl}$. The weight of $\mathrm{AgCl}$ is found to be 14.35 g then %$\left(\mathrm{Ag}+{\mathrm{HNO}}_3\stackrel{\mathrm{NaCl}}{⟶}\mathrm{AgCl}\right)$ of silver in coin is :

Again weightof AgCl given = 14.35 g

Molar mass of AgCl = (103+35.5)g = 143.5 g 

So moles of AgCl $=\frac{14.35}{143.5}\mathrm{mol}$

= 0.1 mol

Applying POAC on Ag,

moles of Ag+ ion present = 0.1 mol

We have molar mass of Ag = 10.8 g Ag

0.1 mol Ag = $108\times 0.1 \mathrm{g} \mathrm{Ag}=10.8 \mathrm{g} \mathrm{Ag}$

Given, mass of the silver coin = 21.6 g

%Ag in the coin = $\frac{\frac{14.35}{143.5}\times 108}{21.6}$

$=50\mathrm{\%}$