Question

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 (ΔT_f) for all four trials, assuming that the freezing point of pure water is 0^oC. Record these values in Data Table II.

2. Calculate the molality of each of the solutions (using the formula ΔT_f = K_f m, where K_f for water is -1.86^oC/m) using the values for freezing point depression you just determined in #1 above, and some algebra in the equation: ΔT_f = K_fm. Record these values in Data Table II.

3. Use these molalities to calculate the moles of solute (moles of ethanol and moles of isopropyl alcohol- remember that molality, m, is moles of solute/kg of solvent). Record these values in Data Table II.

4. The molecular formula for ethyl rubbing alcohol (ethanol) is C₂H₅OH (C₂H₆O) and for isopropyl alcohol is CH₃CHOHCH₃ (C₃H₈O). Calculate the mass of each alcohol in the solution (using the moles of each alcohol calculated in # 3, and the molar mass of each alcohol). Record these values in Data Table II.

5. Divide the mass of each alcohol calculated in # 4 by the density of that alcohol (the density of pure ethanol is 0.789 g/ml and that of pure isopropyl alcohol is 0.781 g/ml). The result will be 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 value in 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 POINT DEPRESSION

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

Answer 1

There is evidently a problem with trial 1 of ethanol because you don't report temperatures under -10 ^oC. So you can't work with trial 1 and only with trial 2. The final temperatures that can be used are those when the system is in equilibrium, this is when the temperature doesn't change with time.

For ethyl alcohol trial 2   For isopropyl alcohol trial 1    For isopropyl alcohol trial 1

25 - (-6) = 31 ^oC 23 - (-1) = 24 ^oC 24 - (-1) = 25 ^oC

Molality: solving for molality

m = = 31 oC / 1.86 = 16.66 m 24/1.86= 12.90 m 25/1.86= 13.44 m

moles number

For 0.002 kg of solvent the moles number = molalityx0.002Kg

moles ethyl alcohol = 0.0333 moles isopropyl alcohol = 0.0258 moles isopropyl alcohol = 0.02688

Mass of alcohol

We need first to calculate the molar weight for each alcohol:

ethyl alcohol ( C2H6O) = (2x12.01) + (6x 1.01) + 16 = 46.08 g/mol

Isopropyl alcohol = (C3H8O) =  (3x12.01) + (8x 1.01) + 16 = 60.11 g/mol

mass of ethyl alcohol = 0.0333 x 46.08 g/mol = 1.53 g

mass Iso. Alc. tiral #1 = 0.0258 x 60.11 g/mol = 1.55 g

mass Iso. Alc trial # 2= 0.02688 x 60.11 g/mol = 1.62 g

Volume of alcohol

Vol ethyl Alc =   1.53g / 0.789 g/ml = 1.94 ml

Vol Iso. Alc Trial # 1 = 1.55 g/ 0.781g/ml = 1.98 ml

Vol Iso. Alc Trial # 2 = 1.62 g/0.781 = 2.07 ml

% alcohol in solution

% ethyl Alc = 1.94 ml/2x100 = 97 %

% Iso. Alc Trial # 1 = 1.98 ml/2x100 = 99 %

% Iso. Alc Trial # 2 = 2.07 ml/2x100 = 103.5 %

% error in purity based on manufacturer’s claim

% error in purity ethyl alc. = 97 -70 = 27

% error Iso. Alc Trial # 1 = 99-70 = 29

%  error Iso. Alc Trial # 2 = 103.5 -70 = 33.5.