No of atoms in 1 mol of CaCO3=6.023×10^23
[tex]\\ \sf\bull\longmapsto No\:of\:moles=\dfrac{4.52\times 10^24}{6.023\times 10^23}[/tex]
[tex]\\ \sf\bull\longmapsto No\:of\:moles=0.7mol[/tex]
Molar mass of CaCO3
[tex]\\ \sf\bull\longmapsto 40u+12u+3(16u)[/tex]
[tex]\\ \sf\bull\longmapsto 52u+48u[/tex]
[tex]\\ \sf\bull\longmapsto 100g/mol[/tex]
Now
[tex]\\ \sf\bull\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}[/tex]
[tex]\\ \sf\bull\longmapsto Given\:mass=No\:of\:moles\times Molar\:Mass[/tex]
[tex]\\ \sf\bull\longmapsto Given\:Mass=0.7(100)[/tex]
[tex]\\ \sf\bull\longmapsto Given\:Mass=70g[/tex]
