Question

How many grams of ammonium fluoride, , must be dissolved to prepare 300. mL of a 0.234 M aqueous solution of the salt

Answer 1

24.24 g

Step-by-step explanation:

First we calculate how many moles are there in 300 mL of a 0.234 M solution, using the definition of molarity:

• Molarity = moles / liters

Converting 300 mL to L ⇒ 300 mL / 1000 = 0.300 L

• 0.234 M = moles / 0.300 L

• moles = 0.78 moles

Then we convert 0.78 moles of NH₄F into grams, using its molar mass:

• 0.78 mol * 31.073 g/mol = 24.24 g