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Chapter 10: Problem 7

Using the phase diagram for water and Raoult's law, explain why salt is spread on the roads in winter (even when it is below freezing).

Short Answer

Expert verified
When salt is spread on roads in winter, it dissolves in the ice and snow and forms a saltwater solution that has a lower freezing point than pure water due to the principle of freezing-point depression. The presence of salt in the solution shifts the equilibrium between ice and liquid, lowering the freezing point of the solution as explained by Raoult's law. As a result, the ice and snow on the roads melt at temperatures below the standard freezing point of water, providing a safer surface for traffic and pedestrians.

Step by step solution

01

Understanding the phase diagram for water

The phase diagram of water plots the relationship between temperature and pressure for water in its solid, liquid, and gaseous states. At the phase boundaries between any two states, the states coexist in equilibrium, which means that water is transforming from one state to another. In the context of this exercise, we are particularly interested in the boundary line between the solid and liquid states. #Step 2: The effect of adding a solute on the freezing point#
02

The effect of adding a solute on the freezing point of water

When a solute like salt (NaCl) is added to water, it dissolves and forms a homogeneous mixture or a solution. The presence of dissolved solute particles disrupts the regular structure of solid with respect to the liquid. As a result, a higher energy barrier is required for water to transition from the liquid state to the solid state. This means that the freezing point of the solution will be lower than the pure water. This is known as freezing-point depression. #Step 3: Raoult's law and its application to freezing-point depression#
03

Raoult's law and its application to freezing-point depression

Raoult's law is an important concept in physical chemistry, which states that the vapor pressure of a component of a solution is proportional to its mole fraction in the solution. In the context of spreading salt on the roads in winter, the vapor pressure for the ice in the saltwater solution will be lower than that of pure ice due to the presence of salt dissolved in the water. As a result, the equilibrium between the ice and the liquid phase will shift towards the liquid state, thus lowering the freezing point of the solution. #Step 4: Applying this knowledge to why salt is spread on roads in winter#
04

Applying this knowledge to why salt is spread on roads in winter

When salt is added to the ice and snow on the road, it forms a saltwater solution, which has a lower freezing point than the pure water. This means that it melts the ice at temperatures below the standard freezing point of water due to freezing point depression. As a result, the ice and snow on the roads will melt, thus providing a safer surface for vehicular traffic and pedestrians alike. Even if the temperature is below the freezing point of pure water, the salt can still be effective in melting the ice on the roads by lowering the freezing point of the resulting solution.

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Most popular questions from this chapter

Formic acid (HCO_A) is a monoprotic acid that ionizes only partially in aqueous solutions. A 0.10-M formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of 0.10 \(M\) formic acid.

Which solvent, water or hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right),\) would you choose to dissolve each of the following? a. \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) b. \(\mathrm{CS}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{OH}\) d. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{CH}_{2} \mathrm{OH}\) e.\(\mathrm HCl\) {f .} \( \mathrm{C}_{6} \mathrm{H}_{6}\)

Is molality or molarity dependent on temperature? Explain your answer. Why is molality, and not molarity, used in the equations describing freezing-point depression and boiling point elevation?

Use the following data for three aqueous solutions of \(\mathrm{CaCl}_{2}\) to calculate the apparent value of the van't Hoff factor. $$\begin{array}{lc} \text { Molality } & \text { Freezing-Point Depression }\left(^{\circ} \mathrm{C}\right) \\ \hline 0.0225 & 0.110 \\ 0.0910 & 0.440 \\ 0.278 & 1.330 \end{array}$$

In a coffee-cup calorimeter, \(1.60 \mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) was mixed with \(75.0 \mathrm{g}\) water at an initial temperature \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\) a. Assuming the solution has a heat capacity of \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\text {soln }}\right)\) for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of kJ/mol. b. If the enthalpy of hydration for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(-630 . \mathrm{kJ} / \mathrm{mol}\), calculate the lattice energy of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)
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