Molar Mass Determination by Depression

Topics: Water, Freezing-point depression, Concentration Pages: 5 (663 words) Published: December 3, 2013

Chemistry 121
Experiment 19
Molar Mass Determination y Depression of the Freezing Point Introduction:
The most commonly used liquid is water. In this experiment we study the equilibria that can exist between pure water and an aqueous solution, and ice, the solid form of water. The heat will transfer from a higher temperature to a lower temperature. In order for water to change states of matter, it takes a certain amount of kinetic energy or heat. The shift from ice to water (solid to a liquid) is called the heat of fusion. The shift from water to ice is called the freezing point of water, which the standard is usually 0°C. This is the point in which water and ice are at equilibrium. The freezing point Tf°, the vapor pressure of water and ice must be equal.

If you add a soluble liquid or solid to the equilibrium mixture, the temperature of the ice and the solution will fall until it reaches equilibrium. The vapor pressure of water at 0°C is less than that of a pure liquid. The temperature of the new solution will change the in which it reaches equilibrium or the new freezing point Tf. The new freezing point will be below the freezing point of the pure liquid.

∆Tf is called the freezing point depression, which is the change in freezing point. Tfinal – Tinitial = ∆Tf . The freezing point depression is one of the colligative properties of a solution. Colligative properties include boiling point elevation, osmotic pressure, and vapor pressure lowering.

When considering colligative properties it is easier to work on Molality. Molality is the solute concentration. Molality of A = no. of moles A dissolved/ no. of kg solvent in solution The boiling point elevation, Tb – kbm = ∆Tb , and the freezing point depression Tf – kfm = ∆Tf , uses the concentration. Kb and Kf are characteristics of the solvent used. They use these characteristics to find the molar mass of an unknown substance.

Finding an unknown, finding two different concentrations, then with the molar mass we can associate if the solution is ionic. The ratio of the true molar mass to the value you find is equal to a quantity. If the molar mass is less than an actual value you have a ionic compound. Hoff factor is the apparent percentage dissociation. Procedure:

1. Determine the freezing point of pure water.
2. Water-ice mixture you need to add a known amount of one of you unknowns 3. Estimate the mass of your liquid to add to the mixture.
4. Measure water ice solution
5. Add solid to a water-ice mixture
6. Record mass
7. Make second molar mass measurement for you liquid unknown 8. Determine molar mass
9. Figure out % dissociation
Data:
A.

Freezing point (°C)
Pure liquid (H20)
0.0

B.
Object :
Mass: (grams)
Styrofoam cup
1.72
Beaker
101.12

Liquid

Trial 1
Trial 2
Mass of liquid (grams)
9.99
7.96
Temperature fallen (°C)
-3.9
-3.2
Overall mass (grams)
150.29
146.2

Trial 1
Trial 2
∆TF (°C)
3.9
3.2
Molality
2.10
1.72
Mass unknown 2
9.99
7.96
Mass of water
(final mass – liquid – Styrofoam cup)
138.58
136.52
1 Kg of solution
72.09
58.306
Molar Mass
34.33
33.90

C.
Object
Mass (grams)
Styrofoam cup
1.72

Solid (unknown 2)
Trial 1
Trial 2
Mass of solid (grams)
10.0
5.94
Freezing point (°C)
-3.2
-2.5
Final Mass (grams)
195.85
167.75

Trial 1
Trial 2
∆TF (°C)
3.2
2.5
Molality
1.72
1.34
Mass solid
10
5.94
Mass of water
(final mass – solid – Styrofoam cup)
184.13
160.09
1 kg
54.31
37.10
Molar Mass
31.58
27.69

Calculation for C.
Actual = 74.55
74.55/29.635 = 2.5
i=3.5 = 3.5-1 x 100 = 250%
average molar mass (31.58+27.69/2) x 250% = 75.56
Discussion:
Molar mass is a colligative property. This property allows us to study the change in states of matter and figure out the solution. The molality of a solution gives us the concentration of the unknown. We then use the freezing point depression to establish the molality of the new substance...
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