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"
