Multiple Choice
Which of the following solutions will have the highest concentration of bromide ions?
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13. Solutions
Osmolarity
Multiple Choice
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
Verified step by step guidance1
First, understand the problem: We need to find out how many milligrams of nitride ions are required to prepare a given volume of a barium nitride solution with a specific molarity.
Identify the chemical formula for barium nitride, which is \( \text{Ba}_3\text{N}_2 \). This indicates that each formula unit contains two nitride ions (\( \text{N}^{3-} \)).
Calculate the number of moles of \( \text{Ba}_3\text{N}_2 \) in the solution using the formula \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \). Convert 820 mL to liters by dividing by 1000.
Use the stoichiometry of the compound \( \text{Ba}_3\text{N}_2 \) to determine the moles of nitride ions. Since there are two nitride ions per formula unit, multiply the moles of \( \text{Ba}_3\text{N}_2 \) by 2.
Convert the moles of nitride ions to milligrams. First, find the molar mass of \( \text{N}^{3-} \) (which is approximately 14.01 g/mol), then convert grams to milligrams by multiplying by 1000.
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