Cryoscopic Determination of Molar Mass

Prepared by: Jerlyn Rose Yasay

Introduction

What is Cryoscopy?

study or practice of determining the freezing point of liquids.

Freezing point

Freezing point depression+ solute

Introduction

Baekmann method,

Introduction

Baekmann method,

Cooled

Steady and Slow

Temperature readings

Introduction

Figure 1 Cooling curve of (a) pure solvent and (b) solution

Introduction

=molar mass of the unknown solute, = , = ; = and is the mass of the solute and

solvent, respectively

Introduction

In the experiment, the molar mass of the unknown solid was determined by the freezing point of the benzene and the freezing point depression of the solution when the unknown was added,

Data and Results

Time

(s)

Benzene°C

Unknown°C

0 10.041 12.12520 8.054 10.30640 6.384 8.19560 5.125 6.72380 5.315 4.793

100 5.322 3.876120 5.323 5.285140 5.33 5.284

Time(s)

Benzene°C

Unknown°C

160 5.33 5.284180 5.33 5.274200 5.33 5.264220 5.33 5.145240 - 4.98260 - 4.7280 - 4.559

Table 1 temperature reading for benzene and the solution at an interval of 20 sec

Data and Results

0 100 200 3000

2

4

6

8

10

12

timr,t

tem

pera

ture

, °

C

0 100 200 3000

2

4

6

8

10

12

14

time, t

Tem

pera

ture

, °C

Figure 2 cooling curve (a) pure solvent To= 5.330 °C (b) solution Tf= 5.256 °C

Data and Results

Trial To T ΔTf M Mave

1 5.330

5.256

-0.074

191.3

195.32 5.36

05.28

4-

0.071199.4

Table 2 Freezing point depression and molar mass of unknown solid for each trial

Discussion

No of solutes=

Freezing point

entropy

Discussion

When a known mass of solute is added in a solvent, the freezing point of the solvent decreases to an extent equal to the number of solute added.

T1

T1

T2 Molar mass

T2 Molar mass

Sample Calculation

Conclusion

color-= dark yellow

Trial To T ΔTf M Mave

1 5.330

5.256

-0.074

191.3

195.32 5.36

05.28

4-

0.071199.4

Conclusion The molar mass of the unknown solid

was 195.3 g/mol .

.;. It is potassium chromate K2CrO4 with 194.19 g/mol.

%error= 0.588%