At 7000 feet above sea level, at what temperature (+/- 2 degrees Fahrenheit) will H2O boil? Assume that the ba

Answer 1

At 7000 feet water will boil at 199.3 F

Boiling Point Elevation
Ethylene Glycol is an organic liquid called anti-freeze which is added to water to make an aqueous solution of Ethylene Glycol and water. This is to prevent water in the radiator of the vehicle from boiling over. It elevates the boiling point of water. At the same time, anti-freeze will depress the freezing point of that same water to prevent freeze up in the winter, hence the name of the solution, "anti-freeze". This ability for a solute to elevate the boiling point and depress the freezing point of the solvent is the focus of this page.

A Molecular View of Boiling Point Variation
Boiling point can be defined in terms of the vapor pressure of the solvent. It is the temperature at which the vapor pressure of the liquid or solvent in a solution is equal to the external pressure. Since vapor pressure increases with increasing temperature, increasing teh temperature until the vapor pressure reaches the external pressure will allow the boiling to occur. We have seen in Raoult's Law that increasing the solute in a solution will depress the vapor pressure. This would result in having to increase the temperature even higher so that the depressed vapor pressure might become equal to the external pressure. In other words from a molecular view we might expect the boiling point to be elevated when solute is increased in a solution.

If we conduct a study of several solutions using the same solvent (water) and the same solute ,but each solution is at a different known concentration. If we determine the boiling point of each solution and then proceed to subtract the boiling point of pure water (100 C) we will get a boiling point difference, Delta Tb. When we plot the Delta Tb's on the "y" axis and the concentration on the "x" axis we should get a linear curve that will cross the "y" axis at zero. In other words, the boiling point difference of the boiling point of the solvent in the solution and the boiling point of the pure solvent is linear with the concentration. If we double the concentration then the difference should double. If we apply the slope intercept formula which can be used for any linear relationship:

y = mx + b
where:

y = whatever you plot on the "y" axis. In our study it would be Delta Tb.

m = slope of the curve which in the case of a linear curve will be constant. In this study the slope is the boiling point elevation constant.

x = whatever you plot on the "x" axis which in this study was the concentration

b = y intercept which is the y value when x = 0. In our study the y intercept is zero

If we plug the Delta Tb in for "y" and make the slope the boiling point elevation constant, Kb. The concentration which will be the "x" values and the y intercept which will be zero

The math relationship for this relationship between boiling point change and concentration is as follows:

Delta Tb = Kb m
In our study we used a non-volatile non-electrolyte as our solute. We find that for all non-electrolytes, there is an equal number of moles involved before and after the solution is formed. We say that the i-factor is 1 for all non-electrolytes. For electrolytes the same equation above would look like this:

Delte Tb = Tb - Tb0 = i Kb m
where i would be greater than 1. In other words, the boiling point elevation would be greater for an electrolyte compared to a non-electrolyte of the same concentration. The concentration must be in moles of solute per kilogram of solvent which is called a molal concentration.

Tb = boiling point of the solvent in the solution

Tb0 = boiling point of the pure solvent

Kb = boiling point elevation constant for the given solvent

m = concentration in molal units, moles of solute / kg of solvent

Let's do a problem showing the application of the boiling point elevation equation.

Calculate the boiling point of a solution of 0.0222 m glucose, a non-electrolyte in water. The Kb for water = 0.512 C/m. Tb0 = 100 C.

Plug in the molal concentration given and the Kb and boiling point of the pure solvent.

Tb - 100 = (0.512) ( 0.0222) = .0114

Tb = 100 + .0114 = 100.0114 C

What would be the boiling point of a solution of the same concentration (.0222m) of an electrolyte solute, Ca3(PO4)2. Kb = 0.512 and m = 0.0222. Assume complete dissociation.

Write the dissociation equation for Ca3(PO4)2
Ca3(PO4)2 + H2O -----> 3Ca+2(aq) + 2 PO4-3(aq)

Calculate the i factor
i = Number of total moles after the solution / Number of moles before the solution = sum of the coefficients of rtoducts on right side / coefficient of solute on left side = 3 + 2 / 1 = 5 / 1 = 5

Using the boiling point elevation equation for an electrolyte calculate the boiling point of the solvent in the solution.
Delta Tb = Tb - 100 = i Kb m

Tb - 100 = (5)(0.512)(.0222) = 0.0568

Tb = 100 + .0568 = 100.0568 C

Answer 2

This is the third time for this question in two days.... is this a national quiz ??

Here's a calculator

http://www.biggreenegg.com/boilingPoint.htm