We are given a compound known to be Sodium Bromine (NaBr). This compound has a molecular weight of
Na=23 g/mol and Br=80 g/mol
NaBr=23+80= 103 g/mol
Assuming if we have 1 mol of NaBr, then we have 80 g of Bromine (Br). Calculating its percent composition in the total NaBr compound, we get
%Br= (80/103)*100=77.7%
Therefore, the answer to the question above is letter A.
