How many milligrams of nitride ions are required to prepare 820 mL of 0.330 M Ba₃N₂ solution?

• A. 120 mg
• B. 320 mg
• C. 560 mg
• D. 760 mg
• E. 7600 mg

Answer: (E)

Verified step by step guidance

1. First, understand that the molarity (M) of a solution is defined as the number of moles of solute per liter of solution. Here, the molarity of the Ba3N2 solution is given as 0.330 M.
2. Calculate the number of moles of Ba3N2 required for 820 mL of solution. Convert the volume from milliliters to liters by dividing by 1000: \( 820 \text{ mL} = 0.820 \text{ L} \). Use the formula: \( \text{moles of Ba3N2} = \text{Molarity} \times \text{Volume in liters} \).
3. Determine the number of moles of nitride ions (N^{3-}) in Ba3N2. The chemical formula Ba3N2 indicates that there are 2 moles of nitride ions for every mole of Ba3N2. Therefore, multiply the moles of Ba3N2 by 2 to find the moles of nitride ions.
4. Convert the moles of nitride ions to grams using the molar mass of nitride ions. The molar mass of N^{3-} is approximately 14.01 g/mol. Use the formula: \( \text{mass in grams} = \text{moles} \times \text{molar mass} \).
5. Finally, convert the mass from grams to milligrams by multiplying by 1000, since 1 gram = 1000 milligrams. This will give you the mass of nitride ions required in milligrams.