Data Table I: Freezing Point Depression
Time (in minutes) | Ethyl Alcohol Trial #1 Temp (in oC) | Ethyl alcohol Trial #2 Temp (in oC) | Isopropyl alcohol Trial #1 Temp (in oC) | Isopropyl alcohol Trial #2 Temp (in oC) |
Initial | 24 | 25 | 23 | 24 |
        1:00 | 10 | 5 | 11 | 11 |
2:00 | 02 | 00 | 7 | 7 |
3:00 | -2 | -2 | 5 | 4 |
4:00 | -5 | -4 | 3 | 2 |
5:00 | -6 | -4 | 2 | 1 |
6:00 | -6 | -5 | 1 | 0 |
7:00 | -7 | -6 | 0 | 0 |
8:00 | -9 | -6 | -1 | 0 |
9:00 | -10 | -6 | -1 | -1 |
10:00 | Below -10 | -7 | -1 | -1 |
11:00 | Below -10 | -7 | -1 | -1 |
12:00 | Below -10 | -7 | -1 | -1 |
13:00 | Below -10 | -7 | -1 | -1 |
14:00 | Below -10 | -7 | -1 | -1 |
15:00 | Below -10 | -7 | -1 | -1 |
16:00 | Below -10 | -7 | -1 | -1 |
17:00 | Below -10 | -7 | -1 | -1 |
18:00 | Below -10 | -7 | -1 | -1 |
19:00 | Below -10 | -7 | -1 | -1 |
20:00 | Below -10 | -7 | -1 | -1 |
21:00 | Below -10 | -7 | -1 | -1 |
22:00 | Below -10 | -7 | -1 | -1 |
23:00 | Below -10 | -7 | -1 | -1 |
24:00 | Below -10 | -6 | -1 | -1 |
25:00 | Below -10 | -6 | -1 | -1 |
26:00 | Below -10 | -6 | -1 | -1 |
27:00 | Below -10 | -6 | -1 | -1 |
28:00 | Below -10 | -6 | -1 | -1 |
29:00 | Below -10 | -6 | -1 | -1 |
30:00 | Below -10 | -6 | -1 | -1 |
31:00 | Below -10 | -6 | -1 | -1 |
32:00 | Below -10 | -6 | -1 | -1 |
33:00 | Below -10 | -6 | -1 | -1 |
34:00 | Below -10 | -6 | -1 | -1 |
35:00 | Below -10 | -6 | -1 | -1 |
36:00 | Below -10 | -6 | -1 | -1 |
37:00 | Below -10 | -6 | -1 | -1 |
38:00 | Below -10 | -6 | -1 | -1 |
39:00 | Below -10 | -6 | -1 | -1 |
Calculations:
Show all work, including formulas used, units/substance labels,and evaluate significant figures when reporting your answer.
1.Calculate the freezing point depression (ΔTf) forall four trials, assuming that the freezing point of pure water is0oC. Record these values in Data Table II.
2. Calculate the molality of each of the solutions (using theformula ΔTf = Kf m, where Kf forwater is -1.86oC/m) using the values for freezing pointdepression you just determined in #1 above, and some algebra in theequation: ΔTf = Kfm. Record these values inData Table II.
3. Use these molalities to calculate the moles of solute (molesof ethanol and moles of isopropyl alcohol- remember that molality,m, is moles of solute/kg of solvent). Record these values in DataTable II.
4. The molecular formula for ethyl rubbing alcohol (ethanol) isC2H5OH (C2H6O) and forisopropyl alcohol is CH3CHOHCH3(C3H8O). Calculate the mass of each alcoholin the solution (using the moles of each alcohol calculated in # 3,and the molar mass of each alcohol). Record these values in DataTable II.
5. Divide the mass of each alcohol calculated in # 4 by thedensity of that alcohol (the density of pure ethanol is 0.789 g/mland that of pure isopropyl alcohol is 0.781 g/ml). The result willbe the volume of the alcohol in mL that is present in the solution.Record this in Data Table II.
6. Divide this volume of alcohol by 2.0 mL and multiply by 100.The result is the % of alcohol in the solution. Record this valuein Data Table II.
7. The alcohols you purchased are reported to be 70% pure.Compare the experimentally determined % of alcohol as calculated in# 6 to the accepted value of 70%, as a % error calculation (% error= │accepted value – experimental value│ / accepted value x 100 %).Record these values in Data Table II.
DATA TABLE II: FREEZING POINTDEPRESSION
Calculated /Derived Value | Ethyl alcohol Trial #1 | Ethyl alcohol Trial # 2 | Isopropyl alcohol Trial # 1 | Isopropyl alcohol Trial # 2 |
∆Tf ( C) | | | | |
Molality (moles solute/ kg solvent) | | | | |
Moles solute | | | | |
Mass of alcohol (g) | | | | |
Volume of alcohol in solution (mL) | | | | |
% alcohol in solution | | | | |
% error in purity based on manufacturer’s claim | | | | |
