There are 6.02E23 molecules present in 1 mol Na2O. In 0.5 mol there will be 0.5 x 6.02E23 molecules.
There are three ions present in 1 molecule Na2O; therefore there will be
0.5 x 3 x 6.02E23 ions present om 0.5 mol Na2O.
thanks
There are three ions present in 1 molecule Na2O; therefore there will be
0.5 x 3 x 6.02E23 ions present om 0.5 mol Na2O.
In the problem above that's 0.5mol x 6.02E23 molecules/mol = ?
In the formula Na2O, there are two sodium ions (Na⁺) and one oxygen ion (O²⁻). The subscript numbers indicate the number of ions of each element in the compound.
So, in 1 mol of Na2O, we have 2 moles of Na⁺ ions and 1 mole of O²⁻ ions.
Since you have 0.5 mol of Na2O, we need to multiply the number of ions by 0.5 to get the number of ions present in that amount.
Number of Na⁺ ions = 2 mol * 0.5 = 1 mol
Number of O²⁻ ions = 1 mol * 0.5 = 0.5 mol
Therefore, there are 1 mole of Na⁺ ions and 0.5 mole of O²⁻ ions present in 0.5 mol of sodium oxide.