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Chapter 10: Problem 6

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 from pure water is dropped into a saltwater solution at \(0^{\circ} \mathrm{C}\), it melts due to the difference in freezing points between pure water and saltwater. The saltwater's lower freezing point causes it to take heat from the ice cube for equilibrium, leading to a gradual melting of the ice cube. This process results in a slight decrease in the temperature and concentration of the saltwater solution, as the melting ice cube dilutes the salt content. The final state of the system reaches a new equilibrium at a slightly lower temperature than \(0^{\circ} \mathrm{C}\), with a reduced saltwater concentration.

Step by step solution

01

Identify the freezing point of saltwater and pure water

The freezing point of pure water is \(0^{\circ} \mathrm{C}\). However, the freezing point of saltwater is lower than that, typically around \(-2^{\circ} \mathrm{C}\), depending on the concentration of the salt in the water. This fact is essential to understand the behavior of the ice cube in the saltwater solution.
02

Determine the initial state

Initially, the ice cube is at \(0^{\circ} \mathrm{C}\), while the saltwater solution is also at \(0^{\circ} \mathrm{C}\). However, the saltwater solution has a freezing point lower than \(0^{\circ} \mathrm{C}\), which means it is in a liquid state. The ice cube is in a solid state.
03

Melting process

As the ice cube (made of pure water) is in contact with the saltwater solution, it will start to melt due to the difference in freezing points. Since the saltwater has a lower freezing point, it will take heat from the ice cube to reach equilibrium. As a result, the ice cube will gradually melt into the saltwater solution.
04

Temperature and concentration changes

During the melting process, the temperature of the saltwater solution may slightly decrease due to the heat taken from the melting ice cube. Additionally, the concentration of the salt in the solution may also decrease since the melting ice cube is pure water, causing dilution of the saltwater.
05

Final state

When the ice cube has melted entirely, the system will reach a new equilibrium at a slightly lower temperature than \(0^{\circ} \mathrm{C}\) because of the melting process. Also, the concentration of the saltwater solution will be lower than initially due to dilution caused by the melting ice cube. In conclusion, an ice cube made from pure water, when dropped into a saltwater solution at \(0^{\circ} \mathrm{C}\), will gradually melt due to the difference in the freezing points. This process results in a slight decrease in the temperature and concentration of the saltwater solution.

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Most popular questions from this chapter

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} /\) mol. Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a 10.0 -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C},\) how many grams of water are present in the sample?

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of \(31.57 \%\) C and \(5.30 \%\) H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of \(-5.20^{\circ} \mathrm{C}\) is recorded for a solution made by dissolving \(10.56 \mathrm{g}\) of the compound in \(25.0 \mathrm{g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Calculate the molarity and mole fraction of acetone in a \(1.00-m\) solution of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) .\) (Density of acetone \(=0.788 \mathrm{g} / \mathrm{cm}^{3} ;\) density of ethanol \(=0.789\) \(\mathrm{g} / \mathrm{cm}^{3} .\) ) Assume that the volumes of acetone and ethanol add.

A solution contains \(3.75 \mathrm{g}\) of a nonvolatile pure hydrocarbon in \(95 \mathrm{g}\) acetone. The boiling points of pure acetone and the solution are \(55.95^{\circ} \mathrm{C}\) and \(56.50^{\circ} \mathrm{C},\) respectively. The molal boilingpoint constant of acetone is \(1.71^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} .\) What is the molar mass of the hydrocarbon?

The solubility of benzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) is \(0.34 \mathrm{g} / 100 \mathrm{mL}\) in water at \(25^{\circ} \mathrm{C}\) and is \(10.0 \mathrm{g} / 100 \mathrm{mL}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(25^{\circ} \mathrm{C}\). Rationalize this solubility behavior. (Hint: Benzoic acid forms a dimer in benzene.) Would benzoic acid be more or less soluble in a 0.1-M NaOH solution than it is in water? Explain.
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