Question

Calculate the volume in milliliters of a 1.33 M iron (II) bromide solution that contains 65.0 g of iron bromide, FeBr2 (FWT = 215.65 g/mol).

Answer 1

Answer: 226.6 mL FeBr₂ solution

Step-by-step explanation:

This is a unit conversion problem, and we are given a mass of FeBr₂ in grams, the formula weight of FeBr₂, and the molarity of the FeBr₂ solution. Molarity is in mol/L, so we will convert from g FeBr₂ → mol FeBr₂ → L FeBr₂ solution → mL FeBr₂ solution. I typed it 2 different ways in case the fractions are difficult to read. :)

65.0g FeBr₂ (
`(1 mol FeBr2)/(215.65g FeBr2)`)(
`(L FeBr2 soln)/(1.33mol FeBr2)`)(
`(1000mL FeBr2 soln)/(L FeBr2 soln)`) = 226.6 mL FeBr₂ soln

65.0g FeBr₂(1 mol FeBr₂/215.65g FeBr₂)(L FeBr₂ soln/1.33 mol FeBr₂)(1000 mL/L) = 226.6 mL FeBr₂

If it's asking about significant figures for your final answer, I would go with 227 mL FeBr₂ solution, since two of our given values had 3 significant figures (65.0 and 1.33), and we use the same number of significant figures as the least precise given value, or the value with the least number of decimal places.

I hope this helped! :)

Answer 2

To calculate the volume of the 1.33 M (molar) iron (II) bromide (FeBr2) solution that contains 65.0 grams of iron bromide, you can use the formula:Volume (in mL)=Mass (in grams)Molar Mass (in g/mol)×Concentration (in M)×1000Volume (in mL)=Molar Mass (in g/mol)×Concentration (in M)Mass (in grams)​×1000First, you need to find the number of moles of iron (II) bromide in 65.0 grams using its molar mass:Molar Mass of FeBr2=2(Atomic Mass of Fe+Atomic Mass of Br)Molar Mass of FeBr2=2(Atomic Mass of Fe+Atomic Mass of Br)Molar Mass of FeBr2=2(55.85g/mol+79.90g/mol)=2(135.75g/mol)=271.50g/molMolar Mass of FeBr2=2(55.85g/mol+79.90g/mol)=2(135.75g/mol)=271.50g/molNow, calculate the number of moles:Number of moles of FeBr2=Mass (in grams)Molar Mass (in g/mol)=65.0g271.50g/mol≈0.2396 molNumber of moles of FeBr2=Molar Mass (in g/mol)Mass (in grams)​=271.50g/mol65.0g​≈0.2396molNow, you can calculate the volume in milliliters:Volume (in mL)=0.2396 mol1.33 M×1000 mL/LVolume (in mL)=1.33M0.2396mol​×1000mL/LVolume (in mL)=0.23961.33×1000 mL≈179.85 mLVolume (in mL)=1.330.2396​×1000mL≈179.85mLSo, the volume of the 1.33 M iron (II) bromide solution that contains 65.0 grams of iron bromide is approximately 179.85 milliliters.