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What would be the mass of 9.03 x 1021 molecules of hydrobromic acid?

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Answer:


1.214\ g

Step-by-step explanation:


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


Now,\\We\ know\ that,\\Mass=(No.\ of\ particles)/(Avagadro's\ Constant)*GMM\\Here,\\Mass\ of\ 9.03*10^(21) molecules\ of\ HBr= (9.03*10^(21))/(6.022*10^(23))*81 \approx 1.214\ g

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