Question

design a protocol by preparing a 100 ml homogeneous solution of hack/fecl3 0.095 m of fecl3, 0.125 m of hcl

Answer 1

We know that Molarity , m = number of moles / volume of the solution in L

                                       = ( mass/molar mass of solute ) / volume of the solution in L

For 0.095M FeCl3 solution (Molar mass of FeCl3 = 162.2 g/mol)

Volume of the solution , V = 100 mL = 0.1 L

0.095M = ( mass / 162.2(g/mol) / 0.1L

mass of FeCl3 = 1.541 g

So exactly weigh and transfer 1.541 g of FeCl3 into a standard flask containg little amount of distilled water and made to dissolve and water upto the mark we get the required solution

For 0.125M HCl solution (Molar mass of HCl = 36.5 g/mol)

Volume of the solution , V = 100 mL = 0.1 L

0.125M = ( mass / 36.5(g/mol) / 0.1L

mass of HCl = 0.456 g

So exactly weigh and transfer 0.456 g of HCl into a standard flask containg little amount of distilled water and made to dissolve and water upto the mark we get the required solution