5.72 x 10^22. There are 5.72 x 10^22 Ca^(2+) ions in 7.36 g Ca3(PO4)2.
a) Moles of Ca3(PO4)2
MM of Ca3(PO4)2 = [3x40.08 + 2(30.97 + 4x16.00)] g/mol
= (120.24 + 2x94.97) g/mol = (120.24 + 189.94) g/mol = 310.18 g/mol
Moles of Ca3(PO4)2
= 7.36 g Ca3(PO4)2 x [1 mol Ca3(PO4)2]/[310.18 g Ca3(PO4)2]
= 0.023 73 mol Ca3(PO4)2
b) Formula units (FU) of Ca3(PO4)2
Formula units of Ca3(PO4)2
= 0.023 73 mol Ca3(PO4)2 x [6.022 x 10^23 FU Ca3(PO4)2]/[1 mol Ca3(PO4)2] = 1.429 x 10^22 FU Ca3(PO4)2
c) Ions of Ca
No. of Ca ions
= 1.429 x 10^22 FU Ca3(PO4)2 x [4 Ca^(2+) ions]/[1 FU Ca3(PO4)2]
= 5.72 x 10^22 Ca^(2+) ions