Question

Excess silver nitrate is reacted with 0.1442 g unknown chloride forming a white
precipitate. The mass of the precipitate is 0.1194 g. Calculate the percent by mass of
chloride in the unknown solid

Answer 1

\[Ag NO _{ 3 } + Cl ^{-} \rightarrow Ag Cl _{(s)} + N O _{3}^{-}\]
Molecular weight of \[AgCl\] is =107.9 + 35.5 = 143.4 g

Mass of chloride in the precipitate is =35.5/143.4*0.1194 = 0.0296g

Percent of chloride is = 0.0296/0.1442*100= 20.5%