The concentration or Molarity of bromide ion in the sample solution is 0.077 M.
In the given question,
0.361 g AgBr × (1 mol AgBr/187.77 g AgBr) × (1 mol Br⁻/1 mol AgBr) × (80.0 g Br⁻/1 mol Br⁻) = 0.1536 g Br⁻
0.1536 × (1 mol Br⁻/80.0 g Br⁻) = 0.00192 mol
Molarity = 0.00192 mol/0.025 L = 0.077 M
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