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Chapter 11: 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 added to water on roads in winter, it lowers the freezing point of water by decreasing the mole fraction of the solvent (water) and reducing the vapor pressure according to Raoult's Law (\(P = P^*_A \cdot x_A\)). This prevents the formation of ice and ensures safer driving conditions, even when temperatures are below 0°C.

Step by step solution

01

Understanding the Phase Diagram of Water

A phase diagram is a graphical representation of the equilibrium between different phases of a substance under varying conditions of temperature and pressure. For water, the phases it can exist in are solid (ice), liquid, and vapor. The phase diagram shows the boundaries where phase transitions happen. The diagram has three axes corresponding to the three phases; the horizontal axis corresponds to the temperature, and the vertical axis corresponds to pressure. At atmospheric pressure, we observe that the freezing point of water is 0°C. Below this temperature, water exists in its solid phase (ice), and above it, in its liquid phase.
02

Introducing Raoult's Law

Raoult's Law states that in a solution of a solvent and a non-volatile solute, the vapor pressure of the solvent over the solution is proportional to the mole fraction of the solvent. The law can be expressed mathematically as: \(P = P^*_A \cdot x_A\) Where: - \(P\) is the vapor pressure of the solution - \(P^*_A\) is the vapor pressure of the pure solvent - \(x_A\) is the mole fraction of the solvent in the solution
03

Explaining the Effects of Salt on Water's Freezing Point

When salt is added to water, it dissolves and forms a solution. According to Raoult's law, as the mole fraction of the solvent (water) decreases due to the presence of the solute (salt), the vapor pressure of the water over the solution decreases. To reach the same vapor pressure as in pure water, the solution (saltwater) must be at a lower temperature than the original freezing point of pure water (0°C). Thus, the introduction of salt lowers the freezing point of water, preventing it from turning into ice even when the temperature is below 0°C.
04

Applying the Effects to Road Safety in Winter

When salt is spread on roads in winter, it lowers the freezing point of any water or melting ice on the road surface. This prevents the formation of ice on the roads, which makes driving safer, as ice can create slippery and dangerous conditions. This is why salt is used during winter months to keep roads safe and free from ice, even when temperatures are below freezing.

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Most popular questions from this chapter

An aqueous solution of 10.00 \(\mathrm{g}\) of catalase, an enzyme found in the liver, has a volume of 1.00 \(\mathrm{L}\) at \(27^{\circ} \mathrm{C}\) . The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be 0.745 torr. Calculate the molar mass of catalase.

What mass of sodium oxalate $\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\( is needed to prepare 0.250 \)\mathrm{L}\( of a \)0.100-M$ solution?

An aqueous solution is 1.00\(\% \mathrm{NaCl}\) by mass and has a density of 1.071 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) . The observed osmotic pressure of this solution is 7.83 atm at \(25^{\circ} \mathrm{C}\) . a. What fraction of the moles of NaCl in this solution exist as ion pairs? b. Calculate the freezing point that would be observed for this solution.

The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CsCl}\) is \(0.320^{\circ} \mathrm{C} .\) The freezing-point depression of a \(0.091-\mathrm{m}\) solution of \(\mathrm{CaCl}_{2}\) is $0.440^{\circ} \mathrm{C} .$ In which solution does ion association appear to be greater? Explain.

Liquid A has vapor pressure \(x\) , and liquid \(\mathrm{B}\) has vapor pressure \(y .\) What is the mole fraction of the liquid mixture if the vapor above the solution is \(30 . \% \mathrm{A}\) by moles? $50 . \% \mathrm{A} ? 80 . \% \mathrm{A}\( ? (Calculate in terms of \)x\( and \)y . )$ Liquid A has vapor pressure \(x,\) liquid \(\mathrm{B}\) has vapor pressure y. What is the mole fraction of the vapor above the solution if the liquid mixture is \(30 . \% \mathrm{A}\) by moles? $50 . \% \mathrm{A} ? 80 . \% \mathrm{A}\( ? (Calculate in terms of \)x\( and \)y . )$
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