Answer in units of g
4.9×10^25 atoms (1 mole B/6.023 x10^23)=81.37 moles
81.37 moles(10.81 g/1mole B)=879.571 grams of Boron
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Verified answer
4.9×10^25 atoms (1 mole B/6.023 x10^23)=81.37 moles
81.37 moles(10.81 g/1mole B)=879.571 grams of Boron