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Chapter 13: Problem 77

Both \(\mathrm{NaCl}\) and \(\mathrm{CaCl}_{2}\) are used to melt ice on roads in winter. What advantages do these substances have over sucrose or urea in lowering the freezing point of water?

Short Answer

Expert verified
The advantages that NaCl and CaCl2 have over sucrose or urea in lowering the freezing point of water is that each molecule of NaCl or CaCl2 breaks into multiple particles when dissolved in water, whereas sucrose and urea only dissolve as single molecules. Therefore, NaCl and CaCl2 are more effective at lowering the freezing point of water because the extent of freezing point depression depends on the number of particles the solute contributes to the solution, not the type of particles.

Step by step solution

01

Understanding the Concept of Freezing Point Depression

Freezing point depression is a phenomenon that occurs when the presence of a solute (i.e., a substance dissolved in a solution) lowers the freezing point of the solvent. When a solvent freezes, it forms a distinct crystal structure. However, when a solute is present, it interferes with the formation of this structure, thereby lowering the freezing point of the solution.
02

Comparing the Number of Particles Released in NaCl, CaCl2, Sucrose, and Urea

When dissolved in water, NaCl and CaCl2 break up into ions. In particular, NaCl breaks up into two particles (Na+ and Cl-), while CaCl2 breaks into three particles (Ca2+ and 2 Cl-). Thus, every molecule of NaCl or CaCl2 contributes multiple ions to the solution. Conversely, sucrose and urea do not break up into ions but dissolve as single molecules, contributing only one particle per molecule to the solution.
03

Explaining the Advantage of NaCl and CaCl2 over Sucrose and Urea

Since the freezing point depression largely depends on the amount of dissolved particles, solutions with NaCl and CaCl2 have a lower freezing point than solutions with sucrose or urea due to the fact that each molecule of NaCl or CaCl2 contribute more particles to the solution than each molecule of sucrose or urea. Thus, NaCl and CaCl2 are superior to sucrose and urea for the purpose of melting ice or lowering the freezing point of water due to their ability to break down into more than one particle in solution.

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Most popular questions from this chapter

A \(3.20-\mathrm{g}\) sample of a salt dissolves in \(9.10 \mathrm{~g}\) of water to give a saturated solution at \(25^{\circ} \mathrm{C}\). What is the solubility (in \(\mathrm{g}\) salt/ \(100 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) ) of the salt?

The concentrated sulfuric acid we use in the laboratory is 98.0 percent \(\mathrm{H}_{2} \mathrm{SO}_{4}\) by mass. Calculate the molality and molarity of the acid solution. The density of the solution is \(1.83 \mathrm{~g} / \mathrm{mL}\).

A solution of \(6.85 \mathrm{~g}\) of a carbohydrate in \(100.0 \mathrm{~g}\) of water has a density of \(1.024 \mathrm{~g} / \mathrm{mL}\) and an osmotic pressure of 4.61 atm at \(20.0^{\circ} \mathrm{C}\). Calculate the molar mass of the carbohydrate.

Describe how you would use the freezing-point depression and osmotic pressure measurements to determine the molar mass of a compound. Why is the boiling- point elevation phenomenon normally not used for this purpose?

A protein has been isolated as a salt with the formula \(\mathrm{Na}_{20} \mathrm{P}\) (this notation means that there are \(20 \mathrm{Na}^{+}\) ions associated with a negatively charged protein \(\mathrm{P}^{20-}\) ). The osmotic pressure of a 10.0 -mL solution containing \(0.225 \mathrm{~g}\) of the protein is \(0.257 \mathrm{~atm}\) at \(25.0^{\circ} \mathrm{C}\). (a) Calculate the molar mass of the protein from these data. (b) What is the actual molar mass of the protein?
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