Calculate the concentration in /molL of the chemist's iron(II) bromide solution.

Question

A chemist prepares a solution of iron(II) bromide FeBr2 by measuring out 0.55kg of iron(II) 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(II) bromide solution.

Answer 1

 

 

The  solution is made of 0.55Kg of FeBr2 because the iron is bivalent(II) as solute and 500 mL of water as solvent, filled till the mark of the flask.

 

There is to calculate the concentration of this solution, in mol/L

 

C= n/V= mol/L

 

The data are: 

 

m FeBr2= 0.55 Kg, and which need to be converted in g, for

 

m= 0.55Kg= 550 g (From  Kg to g there is to multiply for 1000 because of 3 position of difference) 

 

V=500mL equivalent to 0.5 L

 

also in this case, in the international system,  From mL to L, instead there is to divide, positions of distance are 3 as well.

 

The problem gives the Volume, there is to  find the number of moles, n.

 

n from the stochiometry of molarity is equal to

 

n= m/M, where m is the mass in g, and M the molecular weight.

 

We have m, we must calculate M for FeBr2

 

The periodic table provides the info

 

M Fe = 55.847 , MBr= 79.904

 

The formula is FeBr2, for there two atoms of Bromide

 

M FeBr2= 55.847+ 2(79.904)=55.847+ 159.08= 215.655 amu 

 

 

n= m/M = 550/215.655= 2.55 moles

 

after rounded off

 

Finally the concentration will be given after converting the volume from mL to L from

 

 

C= n/V= 2.55/0.5= 5.1 mol/L