Question

Consider a sample of calcium carbonate in the form of a cube measuring 5.093cm on each edge . If the sample has a density of 2.71g.ml^-1 . Calculate the number of moles of calcium carbonate .

Answer 1

V=r³=5.093³=132.1 cm³=132.1 ml

m=2.71x132.1=358 g

$M(CaCO_3)=40+12+16x3=100 g/mol$

$n(CaCO_3)=358/100=3.58 mol$