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(c) 1.5 x 1023
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khushi8765:
tell me how to solve this
atomic mass of sodium = 23g this means 23g has no.of atoms = 6.022 × 10 power 23 therefore 58.5g has no.of atoms = 6.022 × 10 power 23 /23 × 58.5 therefore no.of atoms in an unit cell of bcc structure = 2 therefore no.of unit cell which contains no.of atoms = 6.022 × 10 power 23 × 58.5/23×2 try to solve u ill get the answer
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