The ionic compound, (NH4)2SO4, will fully dissolve in water. If 13.01 g is dissolved in water to give a final volume of 546.0 mL, then find the following ion concentrations.

Question

The concentration of the NH4+ is: _____ M
The concentration of the SO42- is: _____ M

Answer 1

13.01 grams (NH4)2SO4 in 546.0 mL
I will do this with figures that don't show the precision you are using so you will need to recalculate using the number of significant figures you want to use.
molar mass (NH4)2SO4 is approximately 132 so you should recalculate that number to four significant figures.
mols (NH4)2SO4 = g/molar mass = about 13.01/132 = about 0.986
M (NH4)2SO4 = mols/L = about 0.986 moles/0.546 L = about 0.180 M.
So (NH4)^+ = 0.180 M x (2 mols NH4^+ ions/1 mol (NH4)2SO4) = ?
(SO4^2-) = 0.180 x (1 mol SO4^2- ions/1 mol (NH4)2SO4 = ?
It may look complicated but look at it logically. If the concn of NH4)2SO4 is 0.180 M, the sulfate is the same since there is 1 mol sulfate ions in the molecule and there is twice the concn for NH4^+ since there are two NH4+ ions/molecule.