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# Freezing Point Depression

Extracts from this document...

Introduction

Level 1 Pharmacy Physicochemical Principles of Pharmacy 315PMY105 Name: Catherine Magee Student number: 13324063 Practical: 3 - Freezing Point Depression Aims � Demonstration of the effect of solutes used in aqueous formulations on the freezing point of water. � Application of the knowledge of the depression of freezing point to the determination of the molecular weight of an unknown sample. Learning Objectives � Development of the laboratory skills related to weighing samples, making solutions and measuring freezing point depression values � To increase the understanding of the colligative properties of electrolytes and non-electrolytes � Development of computer skills in excel. Introduction Depression of freezing point is one of the colligative properties easiest to measure (others are vapour pressure, boiling point elevation, osmotic pressure). Colligative properties depend on the number of particles present in a solution. The equation for depression of freezing point is as follows: T = Kf i m Where: Kf = molal freezing point depression constant (1.86�C/m for water) m = the molality of the solution (moles of solute/kg solvent) i = the number of particles produced per formula unit Because colligative properties depend on the number of particles in solution, a one molal solution of an electrolyte (NaCl) which dissociates in water will lower the freezing point more than a one molal solution of a non-electrolyte (sucrose). ...read more.

Middle

be able to calculate the i values for the solutions we need to use the equation: T = Kf im NaCl: 4 = 1.86 x i x 1 i = 2.15 Sucrose: 2 = 1.86 x i x 1 i = 1.07 Boiling Points Calculations of NaCl and sucrose boiling points were determined using: ?Tb = Kbm Kb = molal elevation constant M = molality if the solution NaCl: ?Tb = 0.51 x 1 ?Tb = 0.51�C Sucrose: ?Tb = 0.51 x 1 ?Tb = 0.51�C Molecular Mass of Unknown A Molecular mass of solute = (Kf x solute in grams) / (T x solvent in kg) Molecular mass of Unknown A = (1.86 x 1) / (4 x 0.01) Molecular mass of Unknown A = 46.5 kg/mol Table 1: Showing the temperature decrease as time increases using pure Lauric Acid Time (s) Temperature (�C) 0 65 30 61 60 55 90 53 120 49 150 46 180 46 210 45 240 45 270 44 300 44 330 44 360 43 390 43 420 42 450 42 480 41 510 39 540 38 570 37 600 36 630 35 Table 2: Showing the temperature decrease to 25�C as the time increased using 1.0g Lauric Acid & 0.1g Benzoic Acid, 1.0g Lauric Acid & 0.2g Benzoic Acid and 1.0g Lauric Acid & 0.2g Benzoic Acid. ...read more.

Conclusion

Accounting for the Experimental Error The theoretical molar mass of benzoic acid is 122.12 kg/mol and the experimental molar mass was determined as 123.5kg/mol. This is a very slight difference as the low percentage error of 1.13% illustrates. This means that the results are reliable; however this error could be due to the 0.2g of benzoic acid having a lower freezing point of 34�C compared to the 0.3g benzoic acid having a freezing point of 36�C which is not expected. This leads to the very likely chance of a procedural error in weighing out the masses of compounds or carrying out the experiment. Effect of a strong acid on the freezing point If a strong acid was used it would've completely dissociated into its ions and therefore behave in a similar manner as the NaCl did on the freezing point of water; ie, it would lower the freezing point. This occurs because the dissociation means there is an increase in the number of solute particles and leads to the depression of freezing point. Conclusion There are therefore 3 things that can affect the freezing point: 1. addition of a solute (eg; sucrose) - the greater the amount the greater the depression of freezing point 2. addition of a strong acid - dissociation of ions increase the number of solute particles which cause the greatest decrease in freezing point (eg; NaCl) 3. ...read more.

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