Question

You need to make an aqueous solution of 0.213 M iron(III) chloride for an experiment in lab, using a 125 mL volumetric flask. How much solid iron(III) chloride should you add?

Answer 1

FeCl₃ molar mass – 162.2 g/mol

We have to prepare 0.213M in 125ml volumetric flask

Molarity(M) = (Wt/M.Wt)*(1000/volume(ml))

M = 0.213 M

M.Wt = 162.2 g/mol

Volume = 125 ml

We have to calculate Wt of FeCl₃

Molarity(M) = (Wt/M.Wt)*(1000/volume(ml))

Wt = (M*M.Wt*Volume)/1000

Wt = (0.213*162.2*125)/1000

Wt = 4.318 gm

4.318gm FeCl₃ is required.

We have to take 4.318gm FeCl₃ in 125ml volumetric flask and first we have to add some little amount of water, dissolve it and add water upto 125 ml mark.