Answer:
[tex]1.214\ g[/tex]
Explanation:
[tex]We\ know\ that,\\No.\ of\ molecules\ of\ Hydrobromic\ acid=9.03*10^{21} HBr\ molecules\\Now,\\We\ know\ that\ Hydrobromic\ acid\ is\ constituted\ by\ 1\ Hydrogen\ molecule\\ and\ 1\ Bromine\ molecule.\\Gram\ Atomic\ Mass\ of\ Bromine \approx 80\ g\\Gram\ Atomic\ Mass\ of\ Hydrogen =1\ g\\Hence,\\The\ Gram\ Molecular\ Mass\ Of\ Hydrobromic\ Acid=1*1+1*80=81\ g\\Avagadro's\ Constant=6.022*10^{23}\ particles[/tex]
[tex]Now,\\We\ know\ that,\\Mass=\frac{No.\ of\ particles}{Avagadro's\ Constant}*GMM\\Here,\\Mass\ of\ 9.03*10^{21} molecules\ of\ HBr= \frac{9.03*10^{21}}{6.022*10^{23}}*81 \approx 1.214\ g[/tex]
