The number of unit cells present in 58.5g of NaCl is​

Answers (1)

We know

molar mass of NaCl  = 58.5

moles = mass/molar mass

          = 58.5 / 58.5 = 1 mol

 

According to Avagadro's constant, 1 mole of any substance = 6.022 × 1023 molecules/atoms/unit cells.

in bcc, each unit cell contains 4 molecules of NaCl

so,  The number of unit cells present in 58.5g of NaCl is​ 

=  6.022 x 1023/4

= 1.5 x 1023

 

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