A chemist prepares a solution of iron(III) bromide FeBr3 by measuring out 0.613kg of iron(III) bromide into a 500.mL volumetric flask and filling the flask to the mark with water.Calculate the concentration in /molL of the chemist's iron(III) bromide solution. Be sure your answer has the correct number of significant digits.
Question
A chemist prepares a solution of iron(III) bromide FeBr3 by measuring out 0.613kg of iron(III) bromide into a 500.mL volumetric flask and filling the flask to the mark with water.Calculate the concentration in /molL of the chemist's iron(III) bromide solution. Be sure your answer has the correct number of significant digits.
Solution
Sure, here are the steps to calculate the concentration of the iron(III) bromide solution:
Step 1: Convert the mass of iron(III) bromide from kg to g. 1 kg = 1000 g, so 0.613 kg = 0.613 * 1000 = 613 g.
Step 2: Calculate the molar mass of iron(III) bromide (FeBr3). The molar mass of Fe is 55.85 g/mol, and the molar mass of Br is 79.90 g/mol. So, the molar mass of FeBr3 = 55.85 g/mol + 3 * 79.90 g/mol = 295.55 g/mol.
Step 3: Calculate the number of moles of iron(III) bromide. Number of moles = mass / molar mass = 613 g / 295.55 g/mol = 2.073 mol.
Step 4: Convert the volume of the solution from mL to L. 1 L = 1000 mL, so 500 mL = 500 / 1000 = 0.5 L.
Step 5: Calculate the concentration of the solution. Concentration = number of moles / volume = 2.073 mol / 0.5 L = 4.146 mol/L.
So, the concentration of the chemist's iron(III) bromide solution is 4.15 mol/L (rounded to three significant digits).
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