Question

How many atoms of phosphorus are in 7.00 mol of copper(II) phosphate?

Answer 1

Answer: The number of phosphorus atoms in given amount of copper(II) phosphate is $8.431\times 10^{24}$

Explanation:

We are given:

Moles of copper(II) phosphate $Cu_3(PO_4)_2$ = 7.00 mol

1 mole of copper(II) phosphate contains 3 moles of copper, 2 moles of phosphorus and 8 moles of oxygen atoms

Moles of phosphorus in copper(II) phosphate = $(2\times 7.00mol$

According to the mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 7.00 moles of copper(II) phosphate will contain = $(2\times 7\times 6.022\times 10^{23}=8.431\times 10^{24}$ number of phosphorus atoms.

Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is $8.431\times 10^{24}$