23.7 g of Ca₃(PO₄)₎₂ contains 9.20 x 10^22 ions of PO₄³⁻
23.7 g of Ca₃(PO₄)₎₂ contains 0.1528 moles of PO₄³⁻ which is equivalent to 9.20 x 10^22 ions of PO₄³⁻
Further Explanation:An AtomAn atom is the smallest particle of an element that can take part in a chemical reaction. For example, sodium atom, Na. A MoleculeA molecule is a substance that is made up of one or more atoms. The atoms in a molecule may be similar or different making a compound. A compound A compound is a substances that contains two or more different atoms that are bonded together. When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.Ca₃(PO₄)₎₂ is an example of a compound that contains three calcium ions and two Phosphate ions.
Therefore;
Ca₃(PO₄)₎₂ ionizes to give out 3 Ca²⁺ ions and 2 PO₄³⁻ ions
Ca₃(PO₄)₎₂ → 3Ca²⁺ + 2PO₄³⁻
Required to determine the number of ions of PO₄³⁻ ions
Step 1: Moles of Ca₃(PO₄)₎₂
Number of moles = Mass /Molar mass
Molar mass of Ca₃(PO₄)₎₂ = 310.1767 g/mol
Thus;
Moles = 23.7 g/310.1767 g/mol
= 0.0764 moles
Step 3: Moles of PO₄³⁻ ions
1 mole of Ca₃(PO₄)₎₂ ionizes to give 2 moles PO₄³⁻ ions
Therefore;
0.0764 moles will produce;
= 0.0764 moles x 2
= 0.0152 moles PO₄³⁻ ions
Step 3: Number of PO₄³⁻ ions
1 mole of PO₄³⁻ ions contains 6.022 x 10^23 ions
Therefore;
0.01528 moles PO₄³⁻ ions contains;
= 0.01528 x 6.022 x 10^23 ions
= 9.20 x 10^22 ions
Keywords: Atoms, ions, Moles, molar mass
Learn more about; Atom: Molecules Compounds Avogadro’s constant: Molecular mass:
Level: High school
Subject: Chemistry
Topic: Moles
Sub-topic: Avogadro's constant
