# Experiment F: a Raoult’s Law Experiment

Topics: Pressure, Vapor pressure, Partial pressure Pages: 9 (2497 words) Published: January 10, 2013
Physical Chemistry

CHM2330

Experiment F- Raoult's Law Experiment

By: Sanah Assaad
Student Number: 5267864
Partner: Jihad Arafa

T.A: Didier

University of Ottawa

March 25, 2010

Objective:
The purpose of this experiment is to study the total vapour pressure of ideal or non-ideal mixtures of two volatile liquids as a function of chemical composition. Introduction:
For ideal mixtures of volatile liquids the vapour pressure of any given mixture may be obtained by applying Raoult's Law to each of the components of the mixture. If, for example, pA* and pB* are the vapour pressures of pure compound A and B respectively, and A and B are their mole fractions in a particular mixture, then the partial vapour pressures of each component over that mixture at a given temperature are: pA = A pA* and pB = B pB*

and the total vapour pressure is given by the sum of these partial vapour pressures: pT = pA + pB.
Liquid mixtures that obey Raoult's Law essentially define an ideal solution. This means that the presence of A in B has no effect on the vapour pressure of B except by diminishing the number of moles of B present in each unit volume. This can only result when the forces between molecules of A and B are essentially the same as those between A molecules themselves, and B molecules themselves. If the attractive forces between A and B are greater than those between A - A and B - B, then both A and B will exhibit partial pressures PA and PB less than those expected from Raoult's Law. This results in a negative deviation (from that calculated with Raoult's Law) in the total pressure. On the other hand, if the attractive forces between A and B are less than those between A - A and B - B, then both A and B will exhibit partial pressures PA and PB greater than those expected from Raoult's Law. This results in a positive deviation in the total pressure. These deviations can be quite small, but for some mixtures are large enough that the total vapour pressure curves pass through a maximum or minimum.

In this experiment, there two mixture solutions, one is ideal (Hexane-Heptane) and the other is non-ideal (Hexane-Ethanol) behaviour. The diagram below represents the behaviour of the mixtures. In addition, the thin lines represent the ideal behaviour, and the curve lines are for non-ideal behaviour. The curves are due to intermolecular forces.

As the curves approach the extremes of the pure components, molecules of the minor component are surrounded by molecules of the major component. Thus the departures from the ideal give information about the interaction of the components. According to guiding questions #3, there are two pairs of volatile liquids exhibiting both ideal and non-ideal behavior. The ideal pair is Hexane-Heptane mixture and the non-ideal pair is Hexane-Ethanol mixture. Dalton's law of partial pressures states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture. yA = PA/Ptot = xAPA*/Ptot (4)

yB = PB/Ptot = xBPB*/Ptot (5)

According to guiding question #7, the total vapor pressure of binary solutions could be like below:

The total vapour pressure and the two partial When a component (the solvent) is nearly vapour pressures of an ideal binary mixture pure, it has a vapour pressure that is are proportional to the mole fractions of the proportional to the mole fraction with a components.slope p*B (Raoult's law).

When it is the minor component (the solute),
its vapour pressure is still proportional to the
mole fraction, but the constant of
proportionality is now KB(Henry's law).

For guiding question #8, the mole percent of hexane in 30mL of a mixture of hexane and heptane of 20% hexane by volume is shown below (assume room temperature which is 25ºC); Volume of hexane = 30mL * 0.2(20%) = 6mL

Mole of hexane = 6mL(volume) * 0.65483g/mL(density) = 0.0456mol...

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