A solution contains both bromide and nitrate and nitrate is to be detected by the brown ring test. Bromide must first be removed as a precipitate without adding any nitrate. Using Ag 108, S 32 and O 16, the suitable reagent and the minimum mass of it needed for $0.0100 mol$ of $Br X^{-}$ are

Answer: C

Silver sulphate precipitates AgBr and adds no nitrate, whereas

AgNO X_{3}

would itself add

NO X_{3} X^{-}

and ruin the test.

Ag X_{2} SO X_{4} + 2 Br X^{-} 2 AgBr + SO X_{4} X^{2 -}

needs

0.0100/2 = 0.00500 mol

Ag X_{2} SO X_{4}

. With

M = 312

, mass

= 0.00500 \times 312 = 1.56 g