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Assuming complete ionisation, the total number of ions furnished by one unit of ferric alum is (Fe2(So4) 3.(NH4)2SO4.24H2O is?
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Calculation of Total Number of Ions in Ferric Alum

Ferric alum is a hydrated double salt composed of iron(III) sulfate, ammonium sulfate, and water. It has the chemical formula Fe2(SO4)3.(NH4)2SO4.24H2O. When it dissolves in water, it dissociates into its constituent ions.

Step 1: Determine the Number of Ions in Each Compound

  • Fe2(SO4)3: Fe3+ and (SO4)2-

  • (NH4)2SO4: 2NH4+ and (SO4)2-

  • 24H2O: 24H+ and 24OH-



Step 2: Calculate the Total Number of Ions

  • Fe2(SO4)3: 2Fe3+ and 3(SO4)2- (total of 5 ions)

  • (NH4)2SO4: 2NH4+ and (SO4)2- (total of 3 ions)

  • 24H2O: 24H+ and 24OH- (total of 48 ions)



Step 3: Add the Total Number of Ions

  • 2Fe3+ + 2NH4+ + 3(SO4)2- + 24H+ + 24OH- = 2Fe3+ + 2NH4+ + 3(SO4)2- + 48H+ + 48OH-



Step 4: Simplify the Equation

  • 2Fe3+ + 2NH4+ + 3(SO4)2- + 48H+ + 48OH- = 2Fe3+ + 2NH4+ + 3(SO4)2- + 48H2O



Therefore, one unit of ferric alum produces a total of 57 ions (2Fe3+, 2NH4+, 3(SO4)2-, and 48H+ or 48OH-).
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Assuming complete ionisation, the total number of ions furnished by one unit of ferric alum is (Fe2(So4) 3.(NH4)2SO4.24H2O is?
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