Question 1

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

Question 2

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

$\\ \sf\bull\longmapsto No\:of\:moles=(4.52* 10^24)/(6.023* 10^23)$

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

Molar mass of CaCO3

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

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

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

Now

$\\ \sf\bull\longmapsto No\:of\;moles=(Given\:Mass)/(Molar\:Mass)$

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

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

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

Please log in or register to add a comment.

Please log in or register to answer this question.