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 …show more content…
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 =
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 …show more content…
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 =