1gm. of silicon contains 'n' no. of atoms. how many of atoms will be there in 0.8gm. of calcium? (si=29,ca=40)
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Mass of Silicon aton = 29amu
i.e mass of 1 mole silicon atom = 29g
use this relation and
you'll get :
6.022 × 10^23 atom = 29g
so,
1g = (6.022 × 10^23)/29
=2.076 × 10^22 atoms.
so 1 gm of sili on would have 2.076× 10^(22) atoms within it.
now,
as mass of 1 ca atom = 40amu,
therefore
1 mole of ca atoms = 40g
using this relation
1g = (6.022 × 10^23)/ 40
then, 0.8 g
= {(6.022 × 10^23)/40} × 0.8
= 12.088 × 10^21 atoms.
i.e mass of 1 mole silicon atom = 29g
use this relation and
you'll get :
6.022 × 10^23 atom = 29g
so,
1g = (6.022 × 10^23)/29
=2.076 × 10^22 atoms.
so 1 gm of sili on would have 2.076× 10^(22) atoms within it.
now,
as mass of 1 ca atom = 40amu,
therefore
1 mole of ca atoms = 40g
using this relation
1g = (6.022 × 10^23)/ 40
then, 0.8 g
= {(6.022 × 10^23)/40} × 0.8
= 12.088 × 10^21 atoms.
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