Final answer:
In 1.0 L of a 1.0 M solution of mercury(I) nitrate (Hg2(NO3)2), dissociation produces 1 mole of Hg2^2+ ions and 2 moles of NO3^- ions, totaling 3.0 moles of ionic species. The correct answer is c) 3.0 moles.
Step-by-step explanation:
To find out how many moles of ionic species are present in 1.0 L of a 1.0 M solution of mercury(I) nitrate (Hg2(NO3)2), you need to consider the dissociation of the compound in water. Mercury(I) nitrate dissociates into one mercury(I) ion (Hg2^2+) and two nitrate ions (NO3^-) for each formula unit:
Hg2(NO3)2 → Hg2^2+ + 2 NO3^-
For a 1.0 M solution, there is 1 mole of Hg2(NO3)2 dissolved in one liter of solution. The dissociation will produce 1 mole of Hg2^2+ ions and 2 moles of NO3^- ions, giving a total of 3.0 moles of ionic species. Therefore, the correct answer is c) 3.0 moles.