A chemist prepares a solution of iron(II) bromide (FeBr2) by measuring out 46. g of iron(II) bromide into a 400. mL volumetric flask
and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's iron(II) bromide solution. Round your answer to 2 significant digits.
Given mass of FeBr2 = 85.5 g
Molar mass of FeBr2 = 215.65
No. of Moles of FeBr2 = Given Mass / Molar mass = 85.5 / 215.65
No. of Moles = 0.3964
Given Volume is 450 mL i.e. 0.450 L
Now Concentration = 0.3964/0.450
Hence Concentration = 0.881 mol/L