Question

Calculate the normal freezing point of a .6837 M aqueous solution of C12H22O11 that has a density of 1.35 g/ml. (C12H22O11 is a nonvolatile, nondissociating solute). The molar freezing point depression constant of water is 1.86 C/m

Answer 1

Ans. Given, [sucrose] = 0.6837 M

                        Density of solution = 1.35 g/ mL

# Let the volume of solution be 1.0 L

Moles of sucrose in 1.0 L solution = 0.6837 mol

Mass of sucrose in solution = Moles x Molar mass

                                    = 0.6837 mol x (342.30008 g/ mol)

                                    = 234.031 g

# Mass of solution = Volume of solution in mL x Density of solution

                                    = 1000.0 mL x (1.35 g/ mL)

                                    = 1350.0 g

Mass of solvent (water) = Mass of solution – Mass of solute (sucrose)

                                    = 1350.0 g – 234.031 g

                                    = 1115.969 g

                                    = 1.115969 kg

Now,

            Molality of solution = Moles of sucrose / Mass of solvent in kg

                                                = 0.6837 mol / 1.115969 kg

                                                = 0.61265 mol/ kg

                                                = 0.61265 m

# Depression in freezing point of the solution is given by-

                        dT_f = i K_f m             - equation 1

            where, i = Van’t Hoff factor. [i = 2 for NaCl         ; I = 1 for sucrose

                        K_f = molal freezing point depression constant of solvent = 1.86⁰C / m

                        m = molality of the solution

                        dT_f = Freezing point of pure solvent – Freezing point of solution

Putting the values in equation 1-

            dT_f = 1 x (1.86⁰C / m) x 0.61265 m

            Hence, dT_f = 1.14⁰C

# Now, using-

dT_f = Freezing point of pure solvent – Freezing point of solution

Or, 1.14⁰C = 0.0⁰C - Freezing point of solution

            Or, Freezing point of solution = 0.0⁰C – 1.14⁰C

            Hence, Freezing point of solution = -1.14⁰C

Therefore, the normal FP of the solution = -1.14⁰C