The bromide impurity in a 2.00 g sample of a me nitrate is precipitated as silver bromide. If 6.40 ml of 0.200 M ${\mathrm{AgNO}}_3$solution is required, what is the mass percentage of bromide in the sample?

$\underset{\begin{array}{l}\text{ Inorganic }\\ \text{ sample }\end{array}}{{\mathrm{AB}}_r}+\mathrm{AgNO}⟶\mathrm{NgBr}\left(\mathrm{s}\right)+{\mathrm{ANO}}_3$

1 mol of $\mathrm{AgBr}$was produced when 1 mol of AgNO3 precipitated 1 mol of ${\mathrm{Br}}^{-}$ ion. 

It takes 6–4 ml of 0.2 M ${\mathrm{AgNO}}_3$to completely precipitate the contaminant $\left({\mathrm{Br}}^{-}\right)$. 

The necessary amount of ${\mathrm{AgNO}}_3$is 0.0064 x 0.2, or 0.00128 mol. 

0.00128 mol of $\mathrm{AgBr}$were produced in total. 

There are 0.00128 moles of ${\mathrm{Br}}^{-}$in the sample. ${\mathrm{Br}}^{-}$ mass in the sample is

 0.00128 x 80, or 0.1024 g, and its mass percentage in the sample is 0.1024 2.00 x 100, or 0.10242.00 x 100, or 5.12%. 

$\frac{6.4\times 0.2}{1000}\times \frac{80}{2}\times 100 = 5.11$