Glacial Acetic Acid Lab Report

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The procedure of the experiment includes two parts:
1. Determination of the Freezing Point of Glacial Acetic Acid: to record the temperature of acetic acid every 30 seconds until it reach 300 seconds (5 minutes).
2. Determination of the Molar Mass of an Unknown Compound: to record the temperature of the mixture between unknown solid and glacial acetic acid. After that, put the ice-water mixture into the tube and record the temperature every 30 seconds.

Pre lab questions:
1. Solute: dissolved substance into liquid to form solution.
Solvent: dissolving medium in a solution.
Solution: homogeneous mixture of two or more substances.
2. Safety rules for this experiment are wearing safety goggle all the time. Also, glacial acetic acid is toxic
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A. NaCl i = 2
Kf = 1.860C/m m = 0.1m
∆Tf = iKfm = 2 x 1.860C/m x 0.1m = 0.3720C
Tf = Tf0 - ∆Tf = 00C – 0.3720C = -0.3720C
B. Glacial acetic acid i = 1
Kf = 1.860C/m m = 0.1m
∆Tf = iKfm = 1 x 1.860C/m x 0.1m = 0.1860C
Tf = Tf0 - ∆Tf = 00C – 0.1860C = -0.1860C
7. A. BaCl2 i = 3
Kb = 0.5120C/m
M = 0.1m
ΔTb = iKbm = 3*0.5120C/m*0.1m = 0.154 0C
Tb = Tb0 + ΔTb = 100 0C + 0.154 0C = 100.154 0C
B. Surcose i = 1
Kf = 0.512 0C/m m = 0.1 m
ΔTb = iKbm = 1*0.5120C/m*0.1m = 0.0512 0C
Tf = Tb0 + ΔTb = 100 0C + 0.0512 0C = 100.0512 0C
8. Supercooling is a cooling liquid below its freezing point without its changing to a solid. Stirring the liquid will help minimize supercooling.
9. Non-volatile is a substance that do not have to measure vapor pressure
Example: glucose and salt (NaCl)
Volatile is substances that have a measurable vapor pressure
Example: tolure, benzene
10. ΔTf = 3.9 0C
Kf = 20.4 0C/m
Mass of solute = 0.49 g
Mass of solvent = 20g m = ΔTf / Kf = 3.9 0C / 20.4 0C/m = 0.191 m
Moles of solute = molality * mass of solvent 0.191 m * 0.02 kg = 0.0382 mol
Molar mass of solute = mass of solute / moles of solute 0.49 g / 0.0382 mol =

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