Question

How many grmas of CaCO3 are present in a sample if there are 4. 52 x 10^24 atoms of carbon in that sample?

Answer 1

No of atoms in 1 mol of CaCO3=6.023×10^23

$\\ \sf\bull\longmapsto No\:of\:moles=\dfrac{4.52\times 10^24}{6.023\times 10^23}$

$\\ \sf\bull\longmapsto No\:of\:moles=0.7mol$

Molar mass of CaCO3

$\\ \sf\bull\longmapsto 40u+12u+3(16u)$

$\\ \sf\bull\longmapsto 52u+48u$

$\\ \sf\bull\longmapsto 100g/mol$

Now

$\\ \sf\bull\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}$

$\\ \sf\bull\longmapsto Given\:mass=No\:of\:moles\times Molar\:Mass$

$\\ \sf\bull\longmapsto Given\:Mass=0.7(100)$

$\\ \sf\bull\longmapsto Given\:Mass=70g$