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.
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.