Question

7. In a sample of calcium phosphate, Caz(PO4)2,
0.432 mole of phosphorous is present, what is
amount of calcium phosphate is present in the
sample if the sample is 100% pure? [Ca = 40,
P = 31, O = 16]
(1) 98 g
(2) 120 g
(3) 67 g
(4) 34 g​

Answer 1

answer : option (3) 67g

explanation : given, in the sample of calcium phosphate, Ca3(PO4)2, 0.432 mole of phosphorus is present.

but 2 mole of phosphorus is present in one mole of calcium phosphate.

so, 0.432 mole of phosphorus is present in 0.432/2 = 0.216 mole of calcium phosphate.

now molecular mass of calcium phosphate = 3 × atomic mass of Ca + 2 × {atomic mass of P + 4 × atomic mass of O}

= 3 × 40 + 2 × (31 + 4 × 16)

= 120 + 2 × (31 + 64)

= 120 + 2 × 95

= 120 + 190

= 310 g

weight of calcium phosphate in 0.216 mol = 0.216 × 310g

= 66.96 ≈ 67g

hence, option (3) is correct choice.