Question

Calculate the volume in milliliters of a 0.674M iron(III) bromide solution that contains 100.g of iron(III) bromide FeBr3. Be sure your answer has the correct number of significant digits.

Answer 1

We can solve given problem in following steps.

Step 1 : Calculation of moles of iron(III) bromide.

Molar mass of FeBr₃ = 55.85 + ( 3 $\times$ 79.91 ) = 215.67 g / mol

We have , no. of moles = Mass / molar mass

$\therefore$ No. of moles of FeBr₃ = 100.0 g / (  215.67 g / mol ) = 0.46367 mol

Step 2 : Calculation of volume of 0.674 M iron(III) bromide solution.

We know that, Molarity = No. of moles of solute / volume of solution in L

$\therefore$ volume of solution in L = No. of moles of solute / Molarity

$\therefore$ volume of 0.674 M iron(III) bromide solution = 0.46367 mol / 0.674 mol / L

= 0.6879 L

= 687.9 ml

ANSWER : 688 ml of 0.674 M iron(III) bromide solution contain 100 g of iron(III) bromide.