What volume of 0.812 m ammonium sulfate contains 7.65 g of ammonium ion?

The ammonium ion is NH4 + which should have a molecular weight of 18g/mol. Then, the amount of ammonium ion in mol should be 7.65g / (18g/mol) = 0.425 mol.

The formula for ammonium sulfate would be (NH4) 2SO4. For every ammonium sulfate, you need 2 ammonium ion. Then the amount of ammonium sulfate in mol should be: 0.425 mol * (1/2) = 0.2125mol

The volume of the solution would be: 0.2125mol / (0.812 mol/liter) = 0.26169l = 261.70 ml