a 0.5231g sample of impure KBr is treated with excess AgNO3 and 0.8145g AgBr are precipited.?

Question

what is the % purity of the sample?

Answer 1

Assume 100% reaction

MW(AgBr)=108+80=188
MW(KBr)=39+80= 119

KBr + AgNO3 -> AgBr + KNO3

so 1 mole AgBr precipitates from 1 mole KBr or
188 g AgBr precipitate from 119 g KBr
0.8145 g AgBr .. .. .. .. .. .. .. .. .. x

x =119*0.8145/188 =0.5156 g

so the purity is 0.5156/0.5231=98.6%