# TSU Molar Mass Determination by Freezing Point Depression Lab Report

Write 2-3 page lab report on Experiment 4.

LABORATORY DATA SHEET
Name
Experiment 4
Date & Section
Colligative Properties: Molar Mass
Determination by Freezing Point Depression
Experimental Data
0.475
09
1. Mass of the naphthalene
Vfinal (mL)
Vdispensed (mL)
Tfreezing (°C)
Vinitial (mL)
Trial
Mass (g)
1
0.47
10.80
76.541 154
76.51 37.25
-4.5
– 4.8
2
0.48
10.74
3
2. Calculate the molar mass of naphthalene based from your experimental data in each of your 3 trials.
Trial 2: 103.4 g/ind
3. Calculate the average molar mass for naphthalene.
(103.55/nd +103.493/no) (03.45 = 103.53/pol
Avg Molar Massi
2
4. Look up the known molar mass for naphthalene and calculate the percent error in your results.
l actual – Theoretical ! | 12887-103.51
actual
– x 100% =
x 100%=19.2% errer
128.17
24
ATE = To solvent- Tesolution 6.6 – Te Solution
2
= 6-6-(-4.5)
= 11.1°C
Trial I
M=
Д)е де
11.1°C
20.40%
M=0.54 ml/kg
Ke
20.4°/
d = 0.71269/ML
Kg Solvent = lod
= 0.00834 Kg Solvent
1000
#nds=mx kg sovent
– 0.544ack -0.00834 Kg Salvat
22
hsh
0.0045nd naphthalene
nolar mass = mass of naphthalene
&nds naphthalene
0.479 naphthalene
0.004 ond nach Arlene
and Tot
Trial ?
ATE = Te solvent -If Schn=6.661-(-4.9) nds=mx Kg Solvent
Ate = 11.4°C
= 6.559 mag 0.60830 Kg
= 0.00464 mol naphthalene
11.4°C
Ke
20.46%
ndar mass=0.489
20.40
0.00464 mol
M=0.559 moly kg
= 103.4 g no
Kgsolvent=vod
( 000
= (10.74 (0.77269/ML)
M =
ATP ATE
=0.00830 kg
Experiment 4:
COLLIGATIVE PROPERTIES:
MOLAR MASS DETERMINATION BY
FREEZING POINT DEPRESSION
Objective:
The purpose of the experiment is to become familiar with colligative properties and use them to
determine the molar mass of a substance.
Background and Introduction:
Solutions are homogeneous mixtures that contain two or more substances. In simple two component (a.k.a.
binary) solutions, the major component is called the solvent and the minor component is called the solute.
Because the solution is composed primarily of solvent, the physical properties of a solution resemble those of
the solvent. Some of these physical properties, called colligative properties, are independent of the nature of
the solute and depend only on the concentration of the solute in the solution.
Colligative properties include freezing point depression, boiling point elevation, vapor pressure lowering,
and the generation of osmotic pressure. Vapor pressure is the pressure of a vapor in thermodynamic
equilibrium with its condensed phases in a closed container. When the vapor pressure of a liquid is equal to
the atmospheric pressure, the liquid boils. All liquids (and solids for that matter) have a tendency to
evaporate into a gaseous form, and all gases have a tendency to condense back to their liquid or solid form.
Experiments have found that dissolution of a non-volatile solute lowers the vapor pressure of the solvent,
which in turn, raises the boiling point and lowers the freezing point of the solution relative to that of the
pure
solvent.
You may be familiar with some of the common, everyday applications of colligative properties. Antifreeze is
added to the coolant (water) in automotive engines to raise the boiling point and lower the freezing point so
the engine can operate optimally across a wider temperature range that when water alone is used as the
coolant. We spread salt on icy steps in the winter to lower the freezing point and melt the ice. These effects
are expressed qualitatively by the colligative property law, which states that the freezing point and boiling
point differ from those of the pure solvent by amounts that are proportional to the molal (containing 1 mole
of solute per kilogram of solvent) concentration of solute.
The relationship between the lowering of the freezing or boiling point and concentration is shown below:
AT= Km
and AT = Kb.m
Where ATf and ATo are the freezing point depression and boiling point elevation amounts, Kpand Ky are
constants specific for a solvent, and m is the molality of the solution. The Kf constant for cyclohexane, the
solvent used in this experiment, is 20.4°C/m.
Calculations
Example 1
What is the freezing point of a solution containing 1.95 g of biphenyl (C12H10) dissolved in 100 g of
cyclohexane if the normal freezing point of cyclohexane is 6.6°C?
Steps:
1. Calculate the molal concentration of the solution
1 mol C12H10
moles of C12H10 = 1.95g C12H10 X
= 0.0127 mol C12H10
154 g C12H10
kg cyclohexane = 100 g X
1 kg
1000 g
= 0.100 kg cyclohexane
0.0127 mol
m =
mol C12H10
kg cyclohexane
0.127 m
0.100 kg
2. Calculate the change in freezing point
ATF = Kg.m=
·m = (20.4 °C/m)(0.127 m) = 2.6°C
Because freezing points of solutions are always lower than that of the pure solvent, we must subtract the
calculated AT from the normal freezing point of the pure solvent. The solution freezing point will be:
Tf(solvent) – ATf = Tf (solution) = 6.6°C -2.6°C = 4.0°C
Example 2
Determine the molar mass of the compound urea if the freezing point of a solution containing 15 g of urea in
100 g of naphthalene is 63.3°C?
Steps:
1. Determine the AT for the solution
If we look up the freezing point for pure naphthalene in standard table we find it is 80.6°C
ATF = Tf(solvent) – Tf(solution) = 80.6°C – 63.3°C = 17.3°C
2. Calculate the molality of the solution
If we look
up the freezing point constant for pure naphthalene in standard tables we find it is 6.9°C/m.
ATF = Ke.m
m =
AT 17.3°C
= 2.5 m
KF 6.9 °C/m m
21
3. Determine the number of moles of solute from the molality and the given mass of solvent
mol solute
m =
kg solvent
mol solute = m kg solvent = 2.5 m x 0.100 kg = 0.25 mol
4. Calculate the molar mass of the solute
g
*Remember: molar mass =
mol
15 g
molar mass =
0.25 mol
= 60 9/mol
THE BURET AND TITRATION PROCEDURES
Burets are used to measure the volume of a liquid reagent required to react with a carefully measured
(liquid or solid) sample of another substance. Burets are calibrated from top to bottom. 50 ml burets
are calibrated in 0.1 ml increment and thus it is possible to estimate volumes to within 0.01 ml.
Familiarize yourself with the parts of a buret, as shown in Figure 4.1. To properly use and read a
1. Turn the stopcock (used to control the rate at which the liquid is dispensed from the buret) to the
open position (parallel with the buret), rinse the buret once with distilled water and then another 2
times with the liquid that will be dispensed.
2. Hold the buret nearly horizontal and rotate it so that the liquid comes in contact with the entire
interior surface. With the stopcock still open, allow the rinse liquid to drain through the tip.
3. Close the stopcock (perpendicular to the buret) and fill the buret with liquid above the calibration
mark. Gently open the stopcock and allow excess liquid to drain through the tip until no air
bubbles remain and the tip is filled with liquid.
4. Carefully read and record the volume of liquid in the calibrated part of the buret. A small piece
of white
paper should be placed behind the buret to make reading the volume easier.
**Note: It is not necessary to begin with a reading of 0.00 ml, since the volume of
the titrant (liquid being dispensed from the buret) is calculated by the
difference of final and initial volumes.
5. Gently open the stopcock and begin to dispense the titrant into the sample in the receiving
ensure thorough mixing. A piece of white paper should be placed under the receiving flask so
that the indicator (color change) can easily be detected near the endpoint of the titration (when
**Note: The inner wall of the receiving flask should be “washed down” with
distilled water so that all of the titrant comes into contact with the
solution being titrated.
22

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