Question

Which of the following represents the molar quantity of mercury(II) nitrate (Hg(NO₃)₂) in a solution containing 53.90 g of the compound?

A) 0.25 moles
B) 0.50 moles
C) 0.75 moles
D) 1.00 moles

Answer 1

To find the molar quantity of mercury(II) nitrate (Hg(NO₃)₂) in a solution with 53.90 g of the compound, you need to calculate the number of moles using the molar mass of Hg(NO₃)₂ and the given mass. The molar quantity is approximately 0.1748 moles.

Step-by-step explanation:

To find the molar quantity of mercury(II) nitrate (Hg(NO₃)₂) in a solution containing 53.90 g of the compound, we need to use the molar mass of Hg(NO₃)₂ and the given mass. The molar mass of Hg(NO₃)₂ is calculated by adding the atomic masses of its constituent atoms: Hg(atomic mass = 200.59 g/mol) and (NO₃)₂(atomic mass = 2 * (14.01 g/mol for N) + 6 * (16.00 g/mol for O) = 15 * 2 + 16 * 6 = 108.01 g/mol). Therefore, the molar mass of Hg(NO₃)₂ is 200.59 + 108.01 = 308.60 g/mol. Using the formula: moles = mass / molar mass, we can calculate the number of moles of Hg(NO₃)₂ in the solution: moles = 53.90 g / 308.60 g/mol = 0.1748 moles. Therefore, the molar quantity of mercury(II) nitrate in the solution is approximately 0.1748 moles.