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* 10^(24)`

Step-by-step 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* 7.00mol`

According to the mole concept:

1 mole of a compound contains
`6.022* 10^(23)` number of particles

So, 7.00 moles of copper(II) phosphate will contain =
`(2* 7* 6.022* 10^(23)=8.431* 10^(24)` number of phosphorus atoms.

Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is
`8.431* 10^(24)`