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This figure contains a photo of a refinery, showing large columnar structures. A diagram of a fractional distillation column used in separating crude oil is also shown. Near the bottom of the column, an arrow pointing into the column shows a point of entry for heated crude oil. The column contains several layers at which different components are removed. At the very bottom, residue materials are removed as indicated by an arrow out of the column. At each successive level, different materials are removed proceeding from the bottom to the top of the column. The materials are fuel oil, followed by diesel oil, kerosene, naptha, gasoline, and refinery gas at the very top. To the right of the column diagram, a double sided arrow is shown that is blue at the top and gradually changes color to red moving downward. The blue top of the arrow is labeled, “small molecules: low boiling point, very volatile, flows easily, ignites easily.” The red bottom of the arrow is labeled, “large molecules: high boiling point, not very volatile, does not flow easily, does not ignite easily.”
Crude oil is a complex mixture that is separated by large-scale fractional distillation to isolate various simpler mixtures.

Depression of the freezing point of a solvent

Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt ( [link] ), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).

This is a photo of damp brick pavement on which a white crystalline material has been spread.
Rock salt (NaCl), calcium chloride (CaCl 2 ), or a mixture of the two are used to melt ice. (credit: modification of work by Eddie Welker)

The decrease in freezing point of a dilute solution compared to that of the pure solvent, Δ T f , is called the freezing point depression    and is directly proportional to the molal concentration of the solute

Δ T f = K f m

where m is the molal concentration of the solute in the solvent and K f is called the freezing point depression constant    (or cryoscopic constant ). Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of K f for several solvents are listed in [link] .

Calculation of the freezing point of a solution

What is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in [link] ?

Solution

Use the equation relating freezing point depression to solute molality to solve this problem in two steps.

This is a diagram with three boxes connected with two arrows pointing to the right. The first box is labeled, “Molality of solution,” followed by an arrow labeled, “1,” pointing to a second box labeled, “Change in freezing point,” followed by an arrow labeled, “2” pointing to a third box labeled, “New freezing point.”
  1. Calculate the change in freezing point.
    Δ T f = K f m = 5.12 ° C m −1 × 0.33 m = 1.7 ° C
  2. Subtract the freezing point change observed from the pure solvent’s freezing point.
    Freezing Temperature = 5.5 ° C 1.7 ° C = 3.8 ° C

Check your learning

What is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?

Answer:

−9.3 °C

Got questions? Get instant answers now!

Colligative properties and de-icing

Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.

Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft ( [link] ).

This figure contains two photos. The first photo is a rear view of a large highway maintenance truck carrying a bright orange de-icer sign. A white material appears to be deposited at the rear of the truck onto the roadway. The second image is of an airplane being sprayed with a solution to remove ice prior to take off.
Freezing point depression is exploited to remove ice from (a) roadways and (b) the control surfaces of aircraft.

Questions & Answers

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When MP₁ becomes negative, TP start to decline. Extuples Suppose that the short-run production function of certain cut-flower firm is given by: Q=4KL-0.6K2 - 0.112 • Where is quantity of cut flower produced, I is labour input and K is fixed capital input (K-5). Determine the average product of lab
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In economics, the contract curve refers to the set of points in an Edgeworth box diagram where both parties involved in a trade cannot be made better off without making one of them worse off. It represents the Pareto efficient allocations of goods between two individuals or entities,
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Suppose a consumer consuming two commodities X and Y has The following utility function u=X0.4 Y0.6. If the price of the X and Y are 2 and 3 respectively and income Constraint is birr 50. A,Calculate quantities of x and y which maximize utility. B,Calculate value of Lagrange multiplier. C,Calculate quantities of X and Y consumed with a given price. D,alculate optimum level of output .
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suppose the production function is given by ( L, K)=L¼K¾.assuming capital is fixed find APL and MPL. consider the following short run production function:Q=6L²-0.4L³ a) find the value of L that maximizes output b)find the value of L that maximizes marginal product
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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