**Molarity**

The molarity of a solution is given by the expression

#color(blue)(bar(ul(|color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "#

or

#M = n/V#

where

#M =# molarity

#n =# number of moles

#V =# volume in litres

The molar mass of #"FeBr"_3# is 295.56 g/mol.

∴ # n = 12.9 color(red)(cancel(color(black)("g FeBr"_3))) × ("1 mol FeBr"_3)/(295.56 color(red)(cancel(color(black)("g FeBr"_3)))) = "0.043 65 mol FeBr"_3#

#V = 375 color(red)(cancel(color(black)("mL"))) × "1 L"/(1000 color(red)(cancel(color(black)("mL")))) = "0.375 L"#

#M = "0.043 65 mol FeBr"_3/"0.375 L" = "0.116 mol/L"#

**Concentration of #"Fe"^"3+"#**

The equation for the solution process is

#"FeBr"_3(s) → "Fe"^"3+""(aq)" + "3Br"^"-""(aq)"#

We see that 1 mol of #"FeBr"_3# gives 1 mol of #"Fe"^"3+"#.

∴ #["Fe"^"3+"] = (0.116 color(red)(cancel(color(black)("mol FeBr"_3))))/"1 L" × ("1 mol Fe"^"3+")/(1 color(red)(cancel(color(black)("mol FeBr"_3)))) = "0.116 mol/L"#

**Concentration of #"Br"^"-"#**

We see from the chemical equation that 1 mol of #"FeBr"_3# gives 3 mol of #"Br"^"-"#.

∴ #["Br"^"-"] = (0.116 color(red)(cancel(color(black)("mol FeBr"_3))))/"1 L" × ("3 mol Br"^"-")/(1 color(red)(cancel(color(black)("mol FeBr"_3)))) = "0.349 mol/L"#