1) Calculate the freezing point depression expected for an aqueous solution made

Question

1) Calculate the freezing point depression expected for an aqueous solution made by mixing 185.0 grams of glycerol, C3H8O3, with 8.00*102 grams of water. Show your work with correct significant figures. Kf for water is 1.86 °C/m.

2) Calculate the molar mass of a substance if the addition of 42.50 grams of the substance to 400.0g of ethyl acetate (Kb= 2.82 °C/m) elevated the boiling point of the ethyl acetate by 2.26 °C. Show your work with correct significant figures.

3) Calculate the molality of a solution made by adding 78.0 grmas potassium nitrate to 350.0 mL of water. Show calculations with correct significant figures.

Explanation / Answer

Ans. 1. Number of moles of glycerol = Mass/ Molar mass

                                                            = 185.0 g/ (92.09 g/mol)

                                                            = 2.0 mol

Mass of solvent (water) = 8.0 x 102 g = 800.0 g = 0.800 kg                  ; [1 kg = 1000 g]

Molality of solution = Moles of solute (glycerol)/ Mass of solvent (water) in kg

                                    = 2.0 mol/ 0.800 kg

                                    = 2.5 mol/ kg

                                    = 2.5 m

Now, using dTf = i Kf m                     - equation 1

            where, i = Van’t Hoff factor = 1 for non-dissociating solutes

                        Kf = molal freezing point depression constant of the solvent

                        m = molality of the solution

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

                                    = freezing point depression

Putting the values in equation 1-

            dTf = 1 x (1.860C/ m) x 2.5 m = 4.650C

Thus, freezing point depression of resultant solution = 4.650C

Ans. 2. Let the molar mass of the substance by Y g/mol.

Number of moles of substance = Mass/ molar mass

                                                = 42.50 g/ (Y g/mol)

                                                = (42.50/Y) mol

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

                                                = (42.50/Y) mol / 0.400 kg

                                                = (106.25/ Y) m

Given, dTb = 2.260C             ; Kb = 2.820C/m

Now, using dTb = i Kb m                    - equation 2

            where, i = Van’t Hoff factor = 1 for non-dissociating solutes

                        Kb = molal boiling point elevation constant of the solvent

                        m = molality of the solution

                        dTf = boiling point elevation

It’s assumed that the substance is non-ionizing.

Putting the values in equation 2-

            2.620C = 1 x (2.820C/m) x (106.25/ Y) m

            Or, 2.62 = 299.625 / Y

            Or, Y = 299.625/ 2.62 = 114.36

Hence, molar mass of the substance = Y g/mol = 114.36 g/mol

Ans. 3. Moles of KNO3 = mass / Molar mass

                                    = 78.0 g/ (101.10 g/ mol)

                                    = 0.7715 mol

Mass of solvent = 350.0 g = 0.350 kg          ; [assuming density of water is 1.000 g/mL]

Now,

Molality of solution = Moles of KNO3/ mass (in kg) of solvent

                                    = 0.7715 mol / 0.350 kg

                                    = 2.20 mol/kg

                                    = 2.2 m

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