Given that the molality of the Iron(III) chloride solution = 1.541 mol /kg water
1.541 mol [tex]FeCl_{3}[/tex]is present per L solution
Calculating mass of [tex]FeCl_{3}[/tex] from 1.541 mol:
[tex]1.541mol FeCl_{3}*\frac{162.20gFeCl_{3}}{1mol FeCl_{3} }= 249.95 g FeCl_{3}[/tex]
Mass of solvent = 1 kg [tex]*\frac{1000g}{1kg}=1000g solvent[/tex]
Total mass of the solution = 1000 g + 249.95 g = 1249.95 g solution
Density of the solution = [tex]\frac{1249.95g}{1000mL}= 1.25g/mL[/tex]
Molarity of the solution = 1.457 mol/L solution
Calculating mass of [tex]FeCl_{3}[/tex] from 1.457 mol:
[tex]1.457mol FeCl_{3}*\frac{162.20gFeCl_{3}}{1mol FeCl_{3} }= 236.3 g FeCl_{3}[/tex]
Volume of the solution = 1 L *[tex]\frac{1000mL}{1L} =1000mL[/tex]
Mass of solution = [tex]1000mL*\frac{1.25g}{mL} =1250 g solution[/tex]
Mass percentage of [tex]FeCl_{3}[/tex]in the solution =[tex]\frac{236.3g FeCl_{3} }{1250 g solution} *100 =18.9 %[/tex]
Therefore density of the solution is 1.25 g/mL
Mass percent of the solution is 18.9 %