Question

A chemist prepares a solution of iron(III) bromide (FeBr) by measuring out 42.5 g of iron (III) bromide into a 100. mI. volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's ironIII) bromide solution. Be sure your answer has the correct number of significant digits.

Answer 1

[AI-generated answer — Bot GPT 3.5]

To calculate the concentration of the iron (III) bromide solution, we first need to determine the molar mass of FeBr3:

Fe = 55.85 g/mol
Br = 79.90 g/mol

Molar mass of FeBr3 = (55.85 g/mol) + 3(79.90 g/mol) = 295.55 g/mol

Next, we need to convert the mass of iron (III) bromide given (42.5 g) into moles:

moles = mass / molar mass
moles = 42.5 g / 295.55 g/mol
moles = 0.1438 mol

Finally, we can calculate the concentration of the solution:

concentration (mol/L) = moles / volume (L)
volume = 100 mL = 0.1 L

concentration = 0.1438 mol / 0.1 L
concentration = 1.438 mol/L

Therefore, the concentration of the chemist's iron (III) bromide solution is 1.438 mol/L. This value should be reported with 4 significant digits as the mass measurement had 3 significant digits.