Answer: The correct answer is option(4).
Explanation:
Molar mass of the aluminium bromide, \(AlBr_3=266.69 g/mol\)
Atomic mass of bromine = 79.90 g/mol
Percentage of Mass of bromine in \(AlBr_3\):
\(\% Br\text{ by mass}=\frac{\text{Atomic mass of Br}\times \text{Number of Br atoms}\times 100}{\text{Molar mass of }AlBr_3}\)
\(\% Br\text{ by mass}=\frac{79.90 g/mol\times 3\times 100}{266.69 g/mol}=89.97\%\approx 90\%\)
Hence, the correct answer is option(4).
