According to the ionic theory, a basic slag of the following composition: CaO=48.9%, MgO=6.7%, MnO=1.6%, FeO=9.0%, Fe203 =3.3%,Si02 =14.4%,?20 =13.3%andAl^ =4.1% 2+ „2+ M„2+ r.2+ ,3+ aM consistsof Ca",Mg",Mn",Fe",FeJ+ ,Si0",P0",AIO^""andfree 44 0 " ions. Calculate the ion fraction of each ion present in the slag. Mole­ cularweightvaluesare CaO=56.1,MgO=40.3,MnO=70.9,FeO=71.8, Fe203 = 159.7 , Si02 = 60 , P205 = 142.0 and A1203 = 101.9.

To calculate the ion fraction of each ion present in the slag, we first need to calculate the moles of each compound in the slag. This can be done by dividing the percentage composition by the molecular weight of each compound.

1. CaO: 48.9% / 56.1 = 0.8716 moles
2. MgO: 6.7% / 40.3 = 0.1663 moles
3. MnO: 1.6% / 70.9 = 0.0226 moles
4. FeO: 9.0% / 71.8 = 0.1253 moles
5. Fe2O3: 3.3% / 159.7 = 0.0207 moles
6. SiO2: 14.4% / 60 = 0.24 moles
7. P2O5: 13.3% / 142.0 = 0.0937 moles
8. Al2O3: 4.1% / 101.9 = 0.0402 moles

Next, we calculate the total moles in the slag by adding up the moles of each compound: 0.8716 + 0.1663 + 0.0226 + 0.1253 + 0.0207 + 0.24 + 0.0937 + 0.0402 = 1.5804 moles

Finally, we calculate the ion fraction of each ion by dividing the moles of each compound by the total moles:

1. Ca2+: 0.8716 / 1.5804 = 0.5514
2. Mg2+: 0.1663 / 1.5804 = 0.1052
3. Mn2+: 0.0226 / 1.5804 = 0.0143
4. Fe2+: 0.1253 / 1.5804 = 0.0793
5. Fe3+: 0.0207 / 1.5804 = 0.0131
6. SiO2: 0.24 / 1.5804 = 0.1518
7. P2O5: 0.0937 / 1.5804 = 0.0593
8. AlO2-: 0.0402 / 1.5804 = 0.0254

Please note that these values are approximations and the actual ion fractions may vary slightly due to rounding errors.