Solution 2 : Colligative properties

COLLIGATIVE PROPERTIES

1. The Vapor Pressure of Solutions Depression = Penurunan Tekanan Uap Jenuh LarutanVapor pressure is pressure when a liquid and its vapor is in the reversible state.

Francois M. Raoult (1830-1901) did the research on this field, and he stated:”The decrease of vapor pressure equals to the ratio of the number mole of solute and total number of moles in the solution”

Solution SBI 2008 8

Colligative Properties(Cont’d)

Solution SBI 2008 9

po - ps

po

n

n + N=

po = vapor pressure of pure liquidps = vapor pressure of solutionn = number of moles of soluteN = number of moles of solvent

This formula can also be expressed as:

p = vapor pressure decrease (po - ps)X = mole fraction of solute

p = po . X


Solution SBI 2008 10

The Raoult law can not be used for electrolyte solution. For electrolyte solution we must consider the factor i which is the ionization factor. The value of i is:i = (1 + (n-1)); i = van’t Hoff factorn = the number of ions produced = degree of ionization

For example: HCl ionized as HCl H+ + Cl-, number of n is 2, if is 1 then; i = (1 + (2-1) 1 ) = 2

For strong electrolyte close to 1 (~1), but weak electrolyte 0 – 1, non electrolyte 0.



- There are three type of electrolytes:1. Strong electrolyte: it can be dissolved perfectly in

water. Examples of this are salts, strong base and acid; such as: NaCl, KCl, MgCl2, ZnSO4, HCl, CuSO4 , H2SO4, NaOH, KOH, Mg(OH)2, HCl etc

2. Weak electrolyte: only part of it dissolves in water. Examples of this are weak base and weak acid, such as: acetic acid (CH3COOH), ammonium hydroxide (NH4OH), cyanide acid (HCN)

3. Non electrolyte: the substance that can not conduct an electric current. Example of this is many of organic compound; such as sugar (C12H22O11), glucose (C6H12O6), benzene (C6H6) etc.



Example: Determine the vapor pressure of sugar solution 34.2% at 20°C (vapor pressure of water is 17.512 mmHg) ! (Mr sugar 342, water=18)

Answer:1. We must find the mole fraction of sugar solutionAssume the mass of solution is 100 g, so the mass

of sugar based on the mass percent is:Sugar mass = (34.2%/100%) x 100 g = 34.2 gMass of water is = 100 g – mass of sugar

= 100 g – 34.2 g = 65.8 g



Calculate mole of each componentMole of sugar= (34.2 g/342 g/mole) x 1 mole =

0.1 moleMole of water= (65.8 g/18 g/mole) x 1 mole = 3.7

molesMole fraction (X) sugar= 0.1/(0.1 + 3.7) = 0.0263So, the decrease of vapor pressure isp = po x Xsugar = 17.512 x 0.0263

= 0.4606 mmHgSo the pressure of solution is: po - p = 17.512 –

0.4606 = 17.0514

mmHg



Question 1.Calculate the vapor pressure of solution of 19.6 g

H2SO4 (Mr 98) and 90 g of water! The pressure of water in this condition is 74.1 mmHg and H2SO4 is 0.89

Answer:Calculate mole of each componentsMole H2SO4 = 19.6/98 = 0.2

Mole H2O = 90/18 = 5

X H2SO4 = 0.2/5.2 = 0.038

H2SO4 2H+ + SO42-



the number of ions = 3; I = (1 + (3-1).0.89) = 2.78

p = po x X x i

= 74.1 x 0.038 x 2.78= 7.923

The pressure of the solution = 74.1 – 7.923 = 66.177



2. Boiling Point Elevation (Kenaikan titik didih)Boiling point of a liquid (Tb) is temperature when

the vapor pressure of the liquid equal to surround pressure.

Raoult formulated this phenomena as: Tb = Kb . m

Tb = Boiling Point Elevation Change

Kb = molal boiling point constant (°C/mole)

m = molalityFor electrolyte solution: Tb = Kb . m . i

The value of Kb depends on the solvent used.



To calculate the boiling point of the solution (T) is then=

The boiling point of the solvent + Tb

Example 1: Calculate the boiling point of solution made of 12.4 g of ethylenglicol (C2H6O2) and 400 g of water! Kb of water 0.52 °C/mole

Answer: calculate the m of the solutionm = (12.4/62) x (1000/400) = 0.5 Raoult formulated this phenomena as:

Tb = Kb . m = 0.52 °C/mole x 0.5 = 0.26°C

So the solution will boil at = 100 + 0.26° = 100.26°C



Question 2:A chemical substance wants to be determined its

molecular relative mass. 10 g of this substance is dissolved in 500 g of water. If Kb of water is 0.52 °C/mole, then the solution boils at 100.1733°C.

Question 3. 10.125 g of HBr (Mr 81 g/mole) is dissolved in 250 g of water. The solution boils at 100.4628°C. If Kb of water is 0.52 °C/mole, determine the of HBr!



3. Freezing Point Depression (Penurunan titik beku)Freezing point (Tf) is temperature when liquid phase and solid phase is in equilibrium.

Raoult formulated this phenomena as: Tf = Kf . m

Tf = Freezing Point Depression Change

Kf = molal feezing point constant (°C/mole)

m = molalityFor electrolyte solution: Tf = Kf . m . i

The value of Kf depends on the solvent used.



To calculate the freezing point of the solution (T) is then= The feezing point of the solvent + Tf

Example 1. Determine a freezing point of solution composed of 45 g of glucose (Mr 180) in 2000 g of water (Kf water is 1.86 °C/mole)

Answer: calculate m of the glucose solutionm = (45/180) x (1000/2000) = 0.125

Wit Raoult formula : Tf = Kf . M = 1.86 x 0.125 = 0.2325 °C

The freezing point is then = 0 – 0.2325 = -.2325°C



Question 45 g of non electrolyte soultion in 1000 g of water

(Kf water is 1.86 °C/mole) freeze at -0.202°C. Calculate the Mr of this non electrolyte.

Question 5. 29.4 g of H3PO4 (assume as 0.78) dissolves in

600 g of water. In what temperature this solution will freeze?

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Solution 2 : Colligative properties