Question

Post-Lab Material Experiment 19 Data and Calculations: Molar Mass Determination by Depression of the Freezing Point 0.3 Name Section A. Measured Freezing Point of Pure Water B. Finding the Freezing Point of a Solution of Liquid Unknown Target mass of solute (Calculated based on the parameters in the instructions) Unknown # Liquid 3.5 Actual mass of solute used Trial Freezing point of solution (observed) -3.0 Mass of solution 116.6 Trial II Freezing point of solution -3.3 Mass of solution 101.3. Calculations: Trial Trial II Freezing point depression Molality of unknown solution, m, molal molal Mass of solution O Mass of solute O Mass of solvent (water) O Moles of solute E

Answer 1

Topt of fure water (To) = 0.3°C B] Unknown Liquid mass solute (a) = 3.59 Trial-1 fept. of solution () = -3°C mass solution = 116.69 tipte depression (ATA) = To-T, = 0.3-(-3)€ = 3.3°C molality of us known (m) - AIF - 3.3°C, KE 1086odml -1.774 m mass solution = 116.69 mass solute (a) = 3.59. 11 Solvent (b) = 116.6-3.5= 113. 19 moles solute (n) = mxb 9 Unknown solid mass unknown a = 71039 Trial-1 topte solution (T) = -1.9°C mass solution = 81.1g. ATS = 0,3-(1.9) = 2:2°c m= 2.204 = 1.183m. 1.66°mt mass solution - Bling mass solute, a = 7.039 mass solvent, b=8111-7-03 = 74.07g N=mxb = 1.183mpl, x 7407xlo Bvb = 0.0876 mol MW = a = 76039 = 80.23g mot 0,0676 mol l cannot be determined without Knowing exact molar mass. (true) Trial-2 f.pt solution (+1)=-26°c ITS = 0.3-(-206) = 3.1°C m= 2.10 = 1.67m 1.8600m mass solution - 144g a= 76039 b=144-7003 = 136.979. n=1.67 moj x 13697X10² kg = 1.744 mol x 113-1X/63 kg 1-3 I In=0.2 mol Imdarmans = a - 3.5 2 Trial-2 117.5g/mol fipt. solution (T) = -3.3°C mass solution ( = 10l-39 STf = 0-(-3.30€ = 3.3°C mo= ATL = 1. 774m mass solution - 101.39 a = 3.59 b= 101.3 - 315=97.89 n= mxb= 11774 mol x 97.8 xlo 3 Kh - KG 14 = 0, 2287 mol MW = a = 7.03g In-011735 mol = 30.74gmolt 0, 2287 mol MW = 3.5 = |20:173 gomor 011735 mol