Correct option is (1) 5.11
\(\underset{Inorganic \\\,\,sample} {AB_r} + AgNO_3 \longrightarrow AgBr(s) + ANO_3\)
∵ 1 mole of Br- ion precipitated by 1 mole of AgNO3 to form 1 mol AgBr.
∵ 6 - 4 ml at 0.2 M AgNO3 is required to complete precipitate the impurity (Br-).
∴ Number of moles of AgNO3 required = 0.0064 x 0.2 = 0.00128 mol.
∴ Number of moles of AgBr formed = 0.00128 mol.
∴ Number of moles of Br- in sample = 0.00128 mol.
∴ Mass of Br- in sample = 0.00128 x 80 = 0.1024 g
∴ Mass percentage of impurity (Br-) = \(\frac{0.1024}{2.00} \times 100 \) = 5.12 %