/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 You drop an ice cube (made from ... [FREE SOLUTION] | 91Ó°ÊÓ

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You drop an ice cube (made from pure water) into a saltwater solution at \(0^{\circ} \mathrm{C}\) . Explain what happens and why.

Short Answer

Expert verified
When an ice cube made of pure water is dropped into a 0°C saltwater solution, it begins to melt since saltwater remains liquid at 0°C, which is below its freezing point. The melting ice cube dilutes the saltwater, raising its freezing point. Depending on the initial salt concentration and the amount of ice added, part of the saltwater solution might freeze due to this process.

Step by step solution

01

Understand the properties of pure water and saltwater

Pure water freezes at 0°C, which means that at or below this temperature, water turns into ice. However, the freezing point of saltwater is lower than that of pure water due to the dissolved salt that disrupts the formation of ice. So, at 0°C, saltwater is still liquid.
02

Analyze the reaction between ice and saltwater at 0°C

When the ice cube, which is at 0°C, is dropped into the saltwater, the saltwater's temperature remains at 0°C because it is below the freezing point of saltwater. The pure ice cube is at its freezing/melting point, so energy from the saltwater is absorbed by the ice as it begins to melt.
03

Explain the result

As the ice absorbs energy from the saltwater, it starts to melt. As it melts, the saltwater solution becomes slightly more diluted with freshwater, which in turn raises the freezing point. At some point, the temperature might be low enough for the dilute saltwater solution to begin freezing, forming more ice.
04

Conclusion

In conclusion, when an ice cube made of pure water is dropped into a saltwater solution at 0°C, it will begin to melt due to the saltwater being above the freezing point of pure water. The melting of the ice cube will slightly dilute the saltwater, raising its freezing point. Depending on the initial concentration of salt in the solution and the amount of ice added, it may cause part of the saltwater solution to freeze as well.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Changes
Phase changes refer to the transition of a substance from one state of matter to another, such as solid, liquid, or gas. In our scenario, we are interested in the phase change from solid (ice) to liquid (water). When an ice cube is introduced into a saltwater solution, it encounters a unique situation where the two substances are initially at the same temperature.
A phase change occurs because ice needs energy in the form of heat, to melt. This energy is absorbed from the surrounding saltwater. As a result, the ice melts to form water. This melting increases the amount of liquid in the solution, altering its composition slightly.
  • The presence of salt in water inhibits the reformation of ice, lowering the freezing point.
  • The system remains dynamic until an equilibrium related to energy exchange is reached.
  • Over time, as more energy is absorbed, the phase change continues.
Understanding these phase changes is crucial for solving and predicting results in freezing point depression experiments.
Solution Chemistry
Solution chemistry involves understanding how substances dissolve and interact at a molecular level. Salt dissolved in water is an example of a solution where salt (the solute) is evenly dispersed in the solvent (water). In the original problem, the saltwater solution has a lower freezing point due to this molecular interaction. This happens because salt ions disrupt the hydrogen bonding network among water molecules, preventing them from easily forming the rigid structure of ice.
When the ice cube is added to the saltwater, it begins to melt, causing a change in the concentration of the saltwater solution. This is because the newly melted water dilutes the salt concentration.
The key concepts connected to solution chemistry here include:
  • The composition of a solution affects its freezing and boiling points.
  • Adding more solvent (in this case, from the melting ice) can change the concentration of solute in the solution, affecting its properties.
  • The van't Hoff factor gives insight into how solute particles affect the physical properties of the solution.
This knowledge helps us understand why the solution behaves the way it does and the conditions under which freezing—or further melting—will occur.
Thermodynamics
Thermodynamics explores the energy changes within physical and chemical processes. In this exercise, it plays a crucial role in explaining how and why energy is exchanged between the ice cube and the saltwater solution. The primary idea is that energy, in the form of heat, moves from a warmer object to a cooler one until thermal equilibrium is achieved.
As the ice cube absorbs energy, it moves toward melting, illustrating the concept of energy transfer. Thermodynamics further explains:
  • Why the ice cube melts despite the solution's temperature being at 0°C: the ice absorbs energy from the saltwater solution, causing a change in its own state.
  • The concept of equilibrium: once the energy balance is reached, no net melting or freezing occurs.
  • How entropy, the measure of disorder, increases as solid ice becomes liquid water, reflecting a more disordered state.
These principles of thermodynamics help explain the process and predict the final states of both the ice and the saltwater solution in this scenario.

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Most popular questions from this chapter

Which solvent, water or carbon tetrachloride, would you choose to dissolve each of the following? a. \(\mathrm{KrF}_{2}\) b. \(\mathrm{SF}_{2}\) c. \(\mathrm{SO}_{2}\) d. \(\mathrm{CO}_{2}\) e. \(M g F_{2}\) f. \(C H_{2} O\) g. \(C H_{2}=C H_{2}\)

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a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing 1.0 \(\mathrm{g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{g} / \mathrm{mol} )\) if the density of the solution is 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

The vapor pressure of a solution containing 53.6 \(\mathrm{g}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in 133.7 \(\mathrm{g}\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 113 torr at \(40^{\circ} \mathrm{C}\) . Calculate the vapor pressure of pure ethanol at \(40^{\circ} \mathrm{C}\) assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.

A 2.00 -g sample of a large biomolecule was dissolved in 15.0 \(\mathrm{g}\) carbon tetrachloride. The boiling point of this solution was determined to be \(77.85^{\circ} \mathrm{C}\) . Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is \(5.03^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol},\) and the boiling point of pure carbon tetrachloride is \(76.50^{\circ} \mathrm{C} .\)

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