Question

How many moles of ionic species are present in 1.0 L of a 1.0 M mercury(I) nitrate (Hg2(NO3)2) solution?

a) 1.0 moles
b) 2.0 moles
c) 3.0 moles
d) 4.0 moles

Answer 1

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.