Question

Determine the number of atoms of O in 92.3 moles of Cr₃(PO₄)₂.

Answer 1

1 mol H2O contains 1 mol O.

0.16 mol H20 × (1 mol O/1 mol H2O) = 0.16 mol O

1 mole atoms = 6.022×10^23 atoms

0.16 mol O × (6.022×10^23 atoms/mol) = 9.6×10^22 atoms O to two significant figures

There are ~ 9.6×10^22 atoms O in 0.16 mol H2O.

Answer 2

Given:

92.3 moles of Cr₃(PO₄)₂ is given.

To find:

The number of atoms of O in 92.3 moles of Cr₃(PO₄)₂.

Solution:

We know that 1mole atom is equivalent to 6.022x10²³ atoms.

• The total no of O atom in one Cr₃(PO₄)₂ compound= 4x2=8
• Now, 1 mole of Cr₃(PO₄)₂ contains Avogadro no. of Cr₃(PO₄)₂ compound= 6.022 x $10^{23}$ Cr₃(PO₄)₂.

• So, 6.022 x 10²³ Cr₃(PO₄)₂ contains 8x 6.022x 10²³ O atoms. As one Cr₃(PO₄)₂ contains 8 oxygen atoms.
• This means 1 mole Cr₃(PO₄)₂ contains 8x 6.022x 10²³ O atoms.
• So, 92.3 moles Cr₃(PO₄)₂ contains = 92.3x 8x 6.022x 10²³= 4446.6448 x 10²³ ≈4.4467x10²⁶ O atoms.

Final answer:

The number of atoms of O in 92.3 moles of Cr₃(PO₄)₂ is 4.4467x10²⁶.