Question

REPORT SHEET EXPERIMENT Freezing Point 19 Depression A. Preparation of experimental apparatus and acid aPreparation of experimental apparatus and freezing point determinati freezing point determination of laurie I. Mass of freezing point depression apparatus 2. Mass of freezing point depression apparatus with lauric acid 8n 519 ang 3. Mass of laurie acid used mass acid ana nd arpo B. Freezing point determination of a laurie acid solution 4. Designation of unknown 5. Mass of added unknown, Exp. 3 6. Mass of added unknown, Exp. Freezing point of lauric acid from Part A, Exp. Freczing point of lauric acid from Part A. Exp.2 Average freezing point of laurie acid Freezing point of lauric acid and unknown from Part B, Exp.3 Freezing point of lauric acid and unknown from Part B. Exp. 03g Exp.3 Exp.3 Exp.4 Exp.4 Solution molality (show calculations) Moles of unknown (show calculation) Exp.3 Exp.4 Molar mass of unknown (show calculation) Exp.3 Exp. 4 Average molar mass Show your calculations for the solution molality, mole of unknown, and molar mass of unknown. Hand in your cooling curves with your report sheet. 249 Copyright © 2015 PEMSam Education, Inc.

Answer 1

Ans. Step 1: determine experimental dT_f

Freezing point of lauric acid = 43.2⁰C

Freezing point when unknown is added to lauric acid = A⁰C

Freezing point depression, dT_f = 43.2⁰C - A⁰C = 2.0⁰C (say, your experimental dT_f is 2.0⁰C, it will simplify our illustration)

Step 2:

Mass of unknown = 1.039 g

Mass of lauric acid (solvent) = 10.0 g = 0.010 kg      ; [Note: Use exact mass for better accuracy]

Let the molar mass of the unknown = B g/mol

Now,

Number of moles of the unknown = mass / molar mass

                                                            = 1.039 g / (B g/mol)

                                                            = (1.039 / B) mol

Molality of solution = Moles of solute / Kg of solvent

                                    = (1.039/ B) mol / 0.010 kg

                                    = (103.9/ B) mol/ kg                                       ; [1 m = 1 mol/ kg]

                                    = (103.9/ B) m

Step 3: Freezing point depression, dT_f is given by-

            dT_f = i K_f m                - equation 1

            where, i = Van’t Hoff factor. [i = 1 for non-electrolyte solute]. Since lauric acid is a non-polar solvent, it’s assumed that the unknown is also non-polar. So, i = 1.

                        K_f = molal freezing point depression constant of solvent = 3.9⁰C / m

                        m = molality of the solution

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

Putting the values in equation 1-

            2.0⁰C = 1 x (3.9⁰C / m) x (103.9/ B) m

            Or, 2.0⁰C = 405.21⁰C / B

            Or, B = 405.21⁰C / 2.0⁰C = 202.61

Therefore, molar mass of the unknown = B g/mol = 202.61 g/mol

Note: 1. Use the exact value of experimentally obtained dTf

            2. Use exact Kf of lauric acid mentioned in your textbook.