Question

4. Freezing Point Depression and Boiling Point Elevation Problems-7.5 pts Calculate the freezing point depression and the boiling point elevation of a solution containing 17.l g of sucrose, CiH:Oi and 200. g of water. What are the actual freezing and boiling points of the solution? New freezing point New boiling point =

Answer 1

Ans. Moles of sucrose = Mass /Molar mass = 17.1 g / (342.30 g/ mol) = 0.050 mol

Mass of solvent (water) = 200.0 g = 0.200 kg

Now,

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

                                                = 0.050 mol / 0.200 kg

                                                = 0.25 m

#1. 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 = 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 0.25 m

            Hence, dT_f = 0.465⁰C

# Now, using-

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

            Or, 0.465⁰C = 0.0⁰C – FP of solution

            Hence, FP of solution = -0.465⁰C

Therefore, new freezing point = -0.465⁰C

#2. Elevation in boiling point of a solution is given by-

dTb = i Kb m            - equation 1

where, i = Van’t Hoff factor

Kb = molal boiling point elevation constant of solvent

m = molality of the solution

dTb = Boiling point of solution - Boiling point of pure solvent

Putting the values in equation 1-

            dT_b = 1 x (0.512⁰C / m) x 0.25 m

            Hence, dT_b = 0.128⁰C

# Now, using-

            dT_b = BP of solution – BP of pure solvent

            Or, BP of solution = 100.0⁰C + 0.128⁰C = 100.128⁰C

Therefore, new boiling point = 100.128⁰C