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.

Answer 1

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 dT_f = i K_f m                     - equation 1

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

                        K_f = molal freezing point depression constant of the solvent

                        m = molality of the solution

                        dT_f = 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 dT_b = i K_b m                    - equation 2

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

                        K_b = molal boiling point elevation constant of the solvent

                        m = molality of the solution

                        dT_f = 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