Given the molality of CaCl2 solution = 3.17 m
That means 3.17 mol [tex]CaCl_{2}[/tex] are present per 1 kg water.
Mass of [tex]CaCl_{2}[/tex] = [tex]3.17 mol CaCl_{2} * \frac{110.98 g}{1 mol} =351.8 g CaCl_{2}[/tex]
Mass of solution = Mass of solvent + Mass of Solute = 1000 g + 351.8 g = 1351.8 g solution.
Density of solution = 1.24 g/mL
Volume of the solution = [tex]1351.8 g * \frac{1mL}{1.24 g} = 1090 mL[/tex]
Molarity is the moles of solute per L of the solution
Calculating molarity of the solution:
[tex]\frac{3.17 mol}{1090mL*\frac{1L}{1000mL} } =2.908 M[/tex]
