Question

The appropriate number of atoms present in 128gm of calcium oxalate is (calcium oxalate =128gm /mol ).

Answer 1

given wt of calcium oxalate = 128 gm

calcium oxalate = CaC2O4 

Mw of calcium oxalate is 40 + 24 + 64 = 128 gm/mole

mole = wt/Mw = 128/128 = 1 mole

mole = N(no. of atom)/NA(avagadro no.) = N /(6.023 x 10^23) 

1 = N /(6.023 x 10^23) 

so N = 6.023 x 10^23

and no. of atom in calcium oxalate is 7 
 so 7 x 6.023 x 10^23 = 4.2 x 10^24


i hope it will help you 
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