CHM 149 Final Exam July 20.pdf

8/14/2019 CHM 149 Final Exam July 20.pdf

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Final Exam/CHM177

Arceo, Chrissie A. July 20, 2013

2009106926

Plotting the given data, we get the graph: (c0=0.038g/L)

Computing for the molecular weight,

Slope=0.0001=B

M2=10,000 g/mol from intercept =1/M2.


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The threshold for an osmotic effect to occur is at concentration above 0.038g/mL

because at this concentration, the solution has a higher pressure than the solvent.There will be complete membrane permeation if there is no pressure difference betweenthe solute and solvent.

Pressure is dependent on concentration, but concentration on the other hand isindependent. The reason as to why the molecular mass is larger than the computedmolecular weight because polymerization occurs along the addition of solute.

The molecular mass of ether is 10,000 and the computed mass of the ether is about

392. There are about molecules that gave rise to the osmoticpressure generated by the more concentrated solution.

Molar volume:


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Radius:

The molecular weight computed is quite far from the actual value, so it does not seemreasonable. As for the volume and radius of the molecule, I think it is reasonablebecause the radius of the octadecanoic molecule is estimated to be 1.75.

Film pressure involved at the end point of the experiment is the Langmuir-Blodgett filmpressure. This causes the monolayer to become unstable and destroy the monolayer.

The sigma value should be larger. When surfactants are absorbed onto a hydrophobicsurface, the polar head groups face into the solution with the tail pointing outward. Inmore hydrophobic surfaces, surfactants may form a bilayer on the solid, causing it to

become more hydrophilic. The dynamic drop radius can be characterized as the dropbegins to spread. The change in pressure across a given surface is proportional to thegravitational pull on the drop.

A drop of stearic acid (SA) solution is equal to 0.005 ml and the concentration of SA is0.200g/L of benzene. The watch glass diameter is 14cm. To determine the number ofdrops needed to form a monolayer, the following data were gathered on the internet:MW of SA = 284.48 g/mol and cross sectional area of SA= 2.1 x 10-15cm2.

Substituting the radius and area will give:


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To determine the avogadros number, the surface area of water, 49 and crosssectional area of SA= 2.1 x 10-15cm2was used. The no. of molecules was equal to theratio of surface area to CSA.

Avogadros number was computed by dividing the no. of molecules to moles of stearicacid.


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The graph shows a Langmuir isotherm, type II-Sigmoid isotherm aka BET isotherm.That is, the Langmuir is adsorption of the monolayer. The BET isotherm on the otherhand is for the multi-layer. The pressure showing at the range of 0.07-0.09 is for BETand the one in the Langmuir is at the range of 0.03-0.4, the left and right, respectively.

The quantity that can be evaluated from the temperature variation of this discontinuity isdirectly proportional to the values of the pressure. The graph shows the value for thepressure, surface area, and temperature.


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Since temperature and pressure are directly proportional, then, we see the plot to havea rise in the temperature of 10to temperature from 0.079-0.099, about 1. The ratio isabout 10%.


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The Langmuir isotherm plot of the given data is@ T= -63.7

@T = 0


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@T = 5.24

@ T= 20.5

Lettingx

orepresent the total concentration of available sites on a given amount of freshsolid substrate, we can define a fractional coverage () as

(1)where xis the concentration of occupied sites. The rate of adsorption vawill beproportional to the concentration of gas or liquid (c) above the surface and the fraction


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of the surface that is not covered (1-), yielding a rate equation

(2)

where kais the rate constant for adsorption. The rate of desorption is simply

proportional to the fraction of the surface that is already occupied, so the rate equationis

(3)

and kdis the rate constant for desorption. Setting equations (2) and (3) equal yields anequilibrium statement that can be written as

(4)The ratio of rate constants in equation (4) is equal to an equilibrium constant(K= ka/kd). Upon substitution and further rearrangement, the fractional coverage isgiven by

(5)

Equation (5) is plotted in figure (1) for an arbitrary value of K, illustrating how the surfacesites become saturated as the concentration rises.

(figure 1)

The magnitude of K quantifies the relative affinity that a given solute has for surfaceadsorption. Like all equilibrium constants, Kis temperature dependent.

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CHM 149 Final Exam July 20.pdf