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  • Level: GCSE
  • Subject: Maths
  • Word count: 2736

freezing point depression method

Extracts from this document...

Introduction

Experiment 5

Title:

Determining molecular weight by freezing point depression method.

Objectives:

-        To determine freezing point of a substance from cooling curve.

  • Study effect of foreign substance content on freezing point of a solvent.
  • Determine molecular weight using freezing point depression.

Theory & Background:

Dissolution of a substance is a solvent suppresses the freezing point of the solution formed. The freezing depression depends on the amount of the material dissolved in the solvent according to the following equation:-

∆T  =  Kf m  =  Kf   x   ____ __mass of material_______

                                        Molecular weight  X  Kg solvent

Where,

∆T  =          freezing point depression

Kf    =freezing point molar constant of solvent

m    =        molal concentration        =   mass of material/molecular weight  x  kg solvent

                                =   No.of moles of solute/kg solvent        

        Naphthalene is a crystalline, aromatic, white, solid hydrocarbon, best known as the primary ingredient of mothballs. Naphthalene is volatile, forming a flammable vapor. Its molecules consist of two fused benzene rings. It is manufactured from coal tar, and converted to phthalic anhydride for the manufacture of plastics, dyes and solvents. It is also used as an antiseptic and insecticide, especially in mothballs. p-Dichlorobenzene is now often used instead of naphthalene as a mothball substitute. Naphthalene easily sublimates at room temperature. The structure of naphthalene is shown on the next page:

image00.png

Apparatus and Materials:

  • Naphthalene
  • Substance X
  • Test tubes
  • Beaker (500cm3)
  • Thermometer
  • Glass rod

Procedures:

...read more.

Middle

690

720

750

780

810

Temperature (oC), Naphthalene

61

58

57

57

56

55

53

52

50

49

48

Temperature (oC), Naphthalene + (0.50 to 0.75g X)

66

65

64

63

62

60

58

57

55

54

52

Temperature (oC), Naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X)

64

63

62

62

61

60

58

57

56

54

50

Table 3: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 510 - 810 seconds

Time (sec)

840

870

900

930

960

990

1020

1050

1080

Temperature (oC), Naphthalene

46

45

45

45

45

45

43

42

41

Temperature (oC), Naphthalene + (0.50 to 0.75g X)

50

48

45

44

44

44

44

42

41

Temperature (oC), Naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X)

49

47

46

43

43

43

43

43

41

Table 4: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 840 - 1080 seconds

Time (sec)

1110

1140

1170

1200

1230

1260

1290

1320

1350

Temperature (oC), Naphthalene

40

39

38

38

37

37

36

36

36

Temperature (oC), Naphthalene + (0.50 to 0.75g X)

40

38

37

37

36

36

35

35

35

Temperature (oC), Naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X)

41

39

37

36

36

35

35

34

34

Table 5: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 1110 - 1350 seconds

* Since the melting point and freezing of naphthalene, naphthalene + (0.50 to 0.75g X)

   and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) are the same, the graph of  

   temperature versus time for the freezing point is plotted.  

Analysis & Calculation:

  1. Calculate molecular weight of substance X in
  1. Naphthalene + (0.50 to 0.75g X)

Using the formula:

                                        Mass of materials

∆T = Kf m =  Kf  x ------------------------------------------

                               Molecular weight  X  Kg solvent

        ∆T = 78oC -77oC

              = 1 oC

        Kf  = 6.8 g/mol

...read more.

Conclusion

kf value can be determined.

kf  = ( ∆Tf  x kg of solvent x molecular weight of solute )

                --------------------------------------------------------------        

                                grams of solute

2)        Compound A, C8H8N2, melts at 80.5oC and compound B, C10H8, melts at         80.6oC. Compare Z which is given is the same as A or B. Suggest a simple        method to determine the identity of Z.

The same method of freezing point depressing method can be carried out to         determine the molecular weight of the substance Z. If the molecular weight is         132, then the substance is a compound Z. If the molecular weight is 128, then the         substance is compound B. From the experiment, the depression of the         temperature, the mass of solute Z (gram) and the mass of solvent (kg) should be         recorded. Then the molecular weight of the substance Z can be calculated by         using the following equation:

Molecular weight of the solute =  kf (grams of solute)

-----------------------

∆Tf (kg of solvent)

Since the melting point of substance A is 80.5°C and B is 80.6°C, a solvent with a         higher melting point should be used. For example, sodium chloride (NaCl) can be         used as NaCl has a high melting point which is 323 °C.

...read more.

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