Physical Chemistry Experiment - Freezing Point Depression PDF

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PHYSICAL CHEMISTRY LABORATORY NOTEBOOKI. TITLE: Cryoscopy: Freezing Point Depression of a solutionII. OBJECTIVES:To determine the freezing point depression and the new freezing point of the solution containing the unknown To determine the van’t Hoff factor of the unknown solutes as well as the solut...

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PHYSICAL CHEMISTRY LABORATORY NOTEBOOK I.

TITLE: Cryoscopy: Freezing Point Depression of a solution

II.

OBJECTIVES: To determine the freezing point depression and the new freezing point of the solution containing the unknown To determine the van’t Hoff factor of the unknown solutes as well as the solutes’ molar masses

III.

METHODOLOGY: The materials used in the experiment are the following: water bath, test tube, thermometer and a stirrer. The solvents used were water and benzene and the solutes were unknown. To start the experiment, an appropriate amount of solvent was selected to fill the test tube by clicking the solvent menu bar and adjusting the solvent mass button. Then, a thermometer was placed inside the test tube and was subjected to a water bath. The power button was turned on to operate the water bath and the freezing point of the solvent was recorded. Once done, the power button was turned off and an amount of solute was added to the solvent. The water bath was again turned on until the freezing point of the solution was reached. The same procedure was done for all the parts and trials of the experiment. Parts A and B were a combination of a polar solvent and a salt. It is noted that the chosen solvents and solutes for part A and B were not similar. Part C on the other hand was a solution of a polar solvent plus a polar compound. Lastly, part D contained a nonpolar solvent and a nonpolar solute. Three trials were done for each part wherein the solvents and solutes had different amounts. The solutes were assumed to be unknown.

Figure 1. Set up of the online simulated cryoscopy experiment (with permission from https://vlab.amrita.edu/?sub=2&brch=190&sim=337&cnt=4)

Figure 2. Filling the test tube with the selected solvent

Figure 3. Adding an amount of the unknown solute to the solvent to create a solution

IV.

RESULTS:

Table 1. Data table for Part A (polar solvent + salt 1)

solvent used mass solvent (g) freezing point solvent (°C) Kf solute used mass solute (g) freezing point solution (°C) Δ Tf van't Hoff factor solute molar mass solute mean molar mass relative error percent relative error

TRIAL 1 water 21.6 g 0 °C -1.86 Sodium chloride 2.31 g - 6.8 °C

TRIAL 2 water 23.8 g 0 °C -1.86 Sodium chloride 2.46 g - 6.572 °C

- 6.8 °C 2 58.50 g/mol

- 6.572 °C 2 58.51 g/mol 58.51 g/mol 1.1978 x 10−3 0.11%

TRIAL 3 water 25.4 g 0 °C -1.86 Sodium chloride 2.54 g - 6.358 °C - 6.358 °C 2 58.51 g/mol

Table 2. Data table for Part B (polar solvent + salt 2)

solvent used mass solvent (g) freezing point solvent (°C) Kf solute used mass solute (g) freezing point solution (°C) ΔT van't Hoff factor solute molar mass solute mean molar mass relative error Percent relative error

TRIAL 1 water 20.8 g 0 °C -1.86 Aluminum chloride 1.38 g - 3.698 °C

TRIAL 2 water 24.6 g 0 °C -1.86 Aluminum chloride 1.62 g - 3.67 °C

TRIAL 3 water 25.4 g 0 °C -1.86 Aluminum chloride 1.77 g - 3.884 °C

- 3.698 °C 4 133.48g/mol

- 3.67 °C 4 133.50 g/mol 133.49 g/mol 3.747 x 10−4 0.037%

- 3.884 °C 4 133.49 g/mol

Table 3. Data table for Part C (polar solvent + polar compound)

solvent used mass solvent (g) freezing point solvent (°C) Kf solute used mass solute (g) freezing point solution (°C) ΔT van't Hoff factor solute molar mass solute mean molar mass relative error percent relative error

TRIAL 1 water 33.7 g 0 °C -1.86 glucose 3.21 g -0.982 °C

TRIAL 2 water 34.5 g 0 °C -1.86 glucose 3.38 g -1.01 °C

TRIAL 3 water 36 g 0 °C -1.86 glucose 3.15 g - 0.902 °C

-0.982 °C 1 180.42 g/mol

-1.01 °C 1 180.42 g/mol 180.42 g/mol 1.47x10-3 0.15%

- 0.902 °C 1 180.43 g/mol

Table 4. Data table for Part D (nonpolar solvent + nonpolar compound)

solvent used mass solvent (g) freezing point solvent (°C) Kf solute used mass solute (g) freezing point solution (°C) ΔT van't Hoff factor solute molar mass solute mean molar mass relative error Percent relative error

V.

CALCULATIONS:

TRIAL 1 benzene 30.7 g 5.5 °C 5.12 camphor 4.92 g 0.342 °C

TRIAL 2 benzene 31 g 5.5 °C 5.12 camphor 4.8 g 0.516 °C

TRIAL 3 benzene 31.5 g 5.5 °C 5.12 camphor 4.77 g 0.626 °C

- 5.158 °C 1 159.08 g/mol

- 4.984 °C 1 159.06 g/mol 159.03 g/mol 0.04467 4.5 %

- 4.874 °C 1 158.94 g/mol

PART A Kf of water = 1.86 (source: Wired Chemist. (n.d.). Retrieved September 03, 2020, from http://www.wiredchemist.com/chemistry/data/molal-freezing-boiling) van't Hoff factor solute = 2 since NaCl dissociates into two (2) ions which is Na+ and ClReal molar mass of NaCl = 58.44 g/mol (theoretical) Trial 1 ● Δ T = T f (solution) - T f (pure solvent) = - 6.8 °C K (g solute)(i) (−1.86℃/m) (2.31 g N aCl)(2) ● Molar mass = (∆T )(f kg = (− 6.8 °C )(0.0216 kg water) = solution)

58.50 g/mol NaCl

Trial 2 ● Δ T = T f (solution) - T f (pure solvent) = - 6.572 °C K f (g solute)(i) ℃/m) (2.46 g NaCl )(2) ● Molar mass = (∆T )(kg solution) = (−(−1.86 = 58.51 g/mol NaCl 6.572 °C )(0.0238 kg water) Trial 3 ● Δ T = T f (solution) - T f (pure solvent) = - 6.358 °C K f (g solute)(i) (−1.86℃/m) (2.54 g NaCl )(2) ● Molar mass = (∆T )(kg solution) = (− 6.358 °C )(0.0254 kg water) = Mean molar mass = Relative error =

58.50+58.51+58.51 g/mol 3

experimental − theoretical theoretical

Percent relative error =

= 58.51  g/mol NaCl g

=

 58.51 g/mol NaCl

g

(58.51 mol −58.44 mol ) 58.44 g/mol

experimental − theoretical theoretical

= 1.1978 x 10−3 g

x 100 =

g

(58.51 mol −58.44 mol ) 58.44 g/mol

x 100 = 0.11%

PART B Kf of water = 1.86 (source: Wired Chemist. (n.d.). Retrieved September 03, 2020, from http://www.wiredchemist.com/chemistry/data/molal-freezing-boiling) van't Hoff factor solute = 4 since AlCl3 dissociates into four (4) ions which is 1 Al+ and 3 ClReal molar mass of AlCl3 = 133.34 g/mol (theoretical) Trial 1 ● Δ T = T f (solution) - T f (pure solvent) = - 3.698 °C K f (g solute)(i) (−1.86℃/m) (1.38 g AlCl3 )(4) ● Molar mass = (∆T )(kg solution) = (− 3.698 °C )(0.0208 kg water) =

133.48 g/mol AlCl3

Trial 2 ● Δ T = T f (solution) - T f (pure solvent) = - 3.67 °C ● Molar mass =

K f (g solute)(i)

(∆T )(kg solution)

=

(−1.86℃/m) (1.62 g AlCl3 )(4) (−3.67 °C )(0.0246 kg water )

=

133.50 g/mol AlCl3

(−1.86℃/m) ( 1.77g AlCl3 ) (4) = (− 3.884 °C )(0.0254 kg water )

133.49 g/mol AlCl3

Trial 3 ● Δ T = T f (solution) - T f (pure solvent) = - 3.884 °C ● Molar mass =

K f (g solute)(i)

(∆T )(kg solution)

=

Mean molar mass = Relative error =

133.48 + 133.50 + 133.49 3

experimental − theoretical theoretical

Percent relative error =

 g/mol AlCl3 = 133.49 g

=

g

(133.49 mol −133.44mol ) 133.44 g/mol

experimental − theoretical theoretical

= 3.747 x 10−4 g

x 100 =

g

(133.49 mol −133.44mol ) 133.44 g/mol

x 100 = 0.037 %

PART C Kf of water = 1.86 (source: Wired Chemist. (n.d.). Retrieved September 03, 2020, from http://www.wiredchemist.com/chemistry/data/molal-freezing-boiling) van't Hoff factor solute = 1 since glucose is a non-ionic compound and does not dissociate into ions Real molar mass of glucose = 180.156 g/mol (theoretical) Trial 1 ● Δ T = T f (solution) - T f (pure solvent) = - 1.035 °C K f (g solute)(i) (−1.86℃/m) (3.21 g glucose)(1) ● Molar mass = (∆T )(kg solution) = (− 0.982 °C )( 0.0337 kg water ) =

180.42 g/mol glucose

Trial 2 ● Δ T = T f (solution) - T f (pure solvent) = - 1.01 °C K f (g solute)(i)

(−1.86℃/m) (3.38 g glucose)(1) (−1.01 °C )( 0.0345 kg water)

=

180.42 g/mol glucose

● Δ T = T f (solution) - T f (pure solvent) = - 0.902 °C K (g solute)(i) (−1.86℃/m) ( 3.15 g glucose)(1) ● Molar mass = (∆T )(f kg = (− 0.902 °C )( 0.036 kg water) = solution)

 180.43 g/mol glucose

● Molar mass =

(∆T )(kg solution)

=

Trial 3

Mean molar mass = Relative error =

180.42 + 180.42 + 180.43 3

experimental − theoretical theoretical

Percent relative error =

=180.42 g/mol g

=

(180.42 −180.156 mol ) 180.156 g/mol

experimental − theoretical theoretical

= 1.47x10-3 g

x 100 =

g

(180.38 mol −180.156 mol ) 180.156 g/mol

x 100 =  0.15 %

PART D Kf of benzene = (source: Wired Chemist. (n.d.). Retrieved September 03, 2020, from http://www.wiredchemist.com/chemistry/data/molal-freezing-boiling) van't Hoff factor solute = 1 since camphor does not dissociate into ions Real molar mass of = 152.23 g/mol (theoretical)

Trial 1 ● Δ T = T f (solution) - T f (pure solvent) = - 5.158 °C K f (g solute)(i) (−5.12℃/m) (4.92 g camphor ) ● Molar mass = (∆T )(kg solution) = (− = 5.158 °C )( 0.0307kg water)

159.08 g/mol camphor

Trial 2 ● Δ T = T f (solution) - T f (pure solvent) = - 4.984 °C K f (g solute)(i) (−5.12℃/m) (4.8 g camphor) ● Molar mass = (∆T )(kg solution) = (−4.984 °C )( 0.031kg water) =

159.06 g/mol camphor

Trial 3 ● Δ T = T f (solution) - T f (pure solvent) = - 4.878 °C K f (g solute)(i) (−5.12℃/m) (4.77 g camphor) ● Molar mass = (∆T )(kg solution) = (− 4.878 °C )( 0.0315kg water) = Mean molar mass = Relative error =

159.08 + 159.06 + 158.94 3

experimental − theoretical theoretical

Percent relative error = VI.

 g/mol camphor = 159.03 g

=

 158.94 g/mol camphor

g

(159.0 mol −152.23 mol ) 152.23 g/mol

experimental − theoretical theoretical

= 0.04467 g

x 100 =

g

(159.02 mol −152.23 mol ) 152.23 g/mol

x 100 =  4.5 %

CONCLUSION Once a solvent becomes a solution when added with a solute, the freezing point of the solution is lower than that freezing point of the solvent alone. This is due to the particles of the solute present in the solution which slows the solvent’s ability to form a lattice. In other words. In order for a solution to freeze, it needs a lower temperature than the normal freezing point of the solvent. The freezing point of the solutions in each trial were recorded and were used to compute for the freezing point depression using the formula Δ T = T f (solution) - T f (pure solvent). Further, the molar masses of each solute used K f (g solute)(i)

was also calculated using the formula molar mass = (∆T )(kg solution) where Kf is the molal freezing point depression constant of the solvent used, ∆T is the difference between the freezing point of the solution and the freezing point of the solvent solvent and the i is the van’t Hoff factor which depends on the number of ions of the solute in the solution....