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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

Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) and hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) form an ideal solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of pentane and hexane are 511 and \(150 .\) torr, respectively. A solution is prepared by mixing \(25 \mathrm{mL} \text { pentane (density, } 0.63 \mathrm{g} / \mathrm{mL})\) with \(45 \mathrm{mL}\) hexane (density, 0.66 g/mL). a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide. $$\begin{array}{|llc|} \hline & \text { Lattice Energy } & \Delta H_{\text {soln }} \\ \hline \mathrm{CaCl}_{2}(s) & -2247 \mathrm{kJ} / \mathrm{mol} & -46 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Cal}_{2}(s) & -2059 \mathrm{kJ} / \mathrm{mol} & -104 \mathrm{kJ} / \mathrm{mol} \\ \hline \end{array}$$ b. Based on your answers to part a, which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-}\), is more strongly attracted to water?

A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right) .\) The freezing point of the solution is \(2.81^{\circ} \mathrm{C} .\) What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\)

Write equations showing the ions present after the following strong electrolytes are dissolved in water. a. \(\mathrm{HNO}_{3}\) b. \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) c. \(A I\left(N O_{3}\right)_{3}\) d. \(\operatorname{SrBr}_{2}\) e. \(\mathrm{KClO}_{4}\) f. \(\mathrm{NH}_{4} \mathrm{Br}\) g. \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) h. \(\mathrm{CuSO}_{4}\) i. \(\quad\) NaOH

Calculate the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.
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