Multiple Choice
Which of the following solutions will have the highest concentration of bromide ions?
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9. 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 that the molarity (M) of a solution is defined as the number of moles of solute per liter of solution. Here, you are given a 0.330 M solution of Ba3N2, which means there are 0.330 moles of Ba3N2 per liter of solution.
Convert the volume of the solution from milliliters to liters. Since 1 L = 1000 mL, 820 mL is equivalent to 0.820 L.
Calculate the number of moles of Ba3N2 in 0.820 L of a 0.330 M solution using the formula: \( \text{moles of Ba3N2} = \text{Molarity} \times \text{Volume in liters} \).
Determine the number of moles of nitride ions (N^{3-}) in the solution. Since each formula unit of Ba3N2 contains 2 nitride ions, multiply the moles of Ba3N2 by 2 to find the moles of nitride ions.
Convert the moles of nitride ions to milligrams. Use the molar mass of the nitride ion (N^{3-}), which is approximately 14.01 g/mol, to convert moles to grams, and then multiply by 1000 to convert grams to milligrams.
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