## Sunday, October 21, 2007

### Study Guide Ch 9. SOLUTIONS

JEE Syllabus

Solutions:
Raoult's law;
Molecular weight determination from lowering of vapor pressure,
Molecular weight determination from elevation of boiling point and
Molecular weight determination from depression of freezing point.
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Main topics in TMH Book chapter

COMPOSITION OF A SOLUTION
IDEAL LIQUID SOLUTION
COLLIGATIVE PROPERTIES
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Raoult's Law

The vapor pressure and composition in equilibrium with a solution can yield valuable information regarding the thermodynamic properties of the liquids involved. Raoult’s law relates the vapor pressure of components to the composition of the solution. The law assumes ideal behavior. It gives a simple picture of the situation just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approache zero, so the behavior of gases approach the behavior of the ideal gas. Raoult’s law is similar in that it assumes that the physical properties of the components are identical. The more similar the components the more their behavior approaches that described by Raoult’s law.

Using the example of a solution of two liquids, A and B, if no other gases are present the total vapor pressure Ptot above the solution is equal to the sum of the vapor pressures of the two components, P(A) and P(B).

Raoult's Law

The vapor pressure and composition in equilibrium with a solution can yield valuable information regarding the thermodynamic properties of the liquids involved. Raoult’s law relates the vapor pressure of components to the composition of the solution. The law assumes ideal behavior. It gives a simple picture of the situation just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approache zero, so the behavior of gases approach the behavior of the ideal gas. Raoult’s law is similar in that it assumes that the physical properties of the components are identical. The more similar the components the more their behavior approaches that described by Raoult’s law.

Using the example of a solution of two liquids, A and B, if no other gases are present the total vapor pressure Ptot above the solution is equal to the sum of the vapor pressures of the two components, PA and PB.

f the two components are very similar, or in the limiting case, differ only in isotopic content, then the vapor pressure of each component will be equal to the vapor pressure of the pure substance Po times the mole fraction in the solution. This is Raoult’s law.

Thus the total pressure above solution of A and B would be

P(total)= P(A) + P(B)

If the two components are very similar, or in the limiting case, differ only in isotopic content, then the vapor pressure of each component will be equal to the vapor pressure of the pure substance Po times the mole fraction in the solution. This is Raoult’s law.

P(i) = P(i)^0*x(i)

Thus the total pressure above solution of A and B would be

P(total)= =P(A)^0*x(i) + P(B)^0*x(i)

refer: http://www.tannerm.com/raoult.htm

Molecular weight determination from depression of freezing point.

The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.

When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.

The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m

ΔTf = depression in freezing point.

Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf

Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.

To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.

weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb

Molality of the solution, m = Wb*1000/Mb*Wa

Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa

From the above equation Mb can be calculated.

Mb = Kf*Wb*1000/Wa*ΔTf

or

Mb = [Kf*Wb*1000]/[ΔTf * Wa]

Example:

Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)

The formula of Mb is available above.

weight of solute be Wb g = 0.643 g

weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C

Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)

Van't Hoff Factor

In 1986, Van't Hoff introduced a factor ot express the extent of association or dissociation of solutes in solution. It is the ratio of the normal and observed molar masses of the solute.

i = Normal molar mass of the solute/Observed molar mass of the solute

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JEE Question 2007 paper I

When 20 g of naphthoic acid (C-11H-8O-2) is dissolved in 50 g of benzene (K-f = 1.72 K kg mol^-1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is

(A) 0.5
(B) 1
(C) 2
(D) 3

Van't Hoff factor = Normal Molar Mass/Observed Molar mass

Observed Molar Mass M-B = K-f*1000*W-B/(W-A*ΔT-f)

K-f = Molal depression constant
W-A = weight of solvent
W-B = weight of solute
ΔT-f = depression in freezing point.

In the problem Normal Molar mass = 172 (12*11+8*1+2*16 = 132+8+32 = 172)
Observe molar mass = 1.72*1000*20/50*2 = 17.2*20 = 172*2

Van't Hoff factor =172/172*2 = 0.5

Ref: Dr. Jauhar's book unit 3: solutions.
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