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Chapter 11: Problem 10

Consider a beaker of salt water sitting open in a room. Over time, does the vapor pressure increase, decrease, or stay the same? Explain.

Short Answer

Expert verified
In conclusion, the vapor pressure of saltwater in an open beaker decreases over time due to the evaporation of water, which leads to an increase in the salinity of the remaining water. This increased salinity lowers the vapor pressure of water according to Raoult's law.

Step by step solution

01

Vapor pressure is the pressure exerted by a vapor in equilibrium with its condensed phases in a closed container. For our saltwater beaker, the vapor in question is water vapor, which evaporates from the liquid saltwater. The rate at which water molecules evaporate into the air depends on the temperature and the concentration of the saltwater. #Step 2: Analyze the effect of evaporation#

Over time, when the saltwater beaker is exposed to the room's atmosphere, water molecules will evaporate from the liquid surface into the air. This evaporation will lead to a decrease in the volume of water in the beaker. However, since the beaker is open to the air, this loss of water vapor will not result in an increase in the vapor pressure. #Step 3: Consider the salinity of the water#
02

As water evaporates from the surface, the concentration of salts in the remaining water will increase. The increase in salt concentration has an effect on the vapor pressure of water: The higher the salinity of the water, the lower the vapor pressure will be compared to pure water. This is a phenomenon known as Raoult's law, which states that the vapor pressure of a component in a mixture is proportional to the mole fraction of that component in the solution. #Step 4: Conclusion#

In conclusion, as the evaporation of water continues and the salinity of the remaining water increases, the vapor pressure of the saltwater will decrease over time.

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Most popular questions from this chapter

A \(1.37 M\) solution of citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) in water has a density of \(1.10 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the mass percent, molality, mole fraction, and normality of the citric acid. Citric acid has three acidic protons.

For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

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

A solution is prepared by mixing \(0.0300 \mathrm{~mol} \mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(0.0500\) \(\mathrm{mol} \mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at \(25^{\circ} \mathrm{C}\). At \(25^{\circ} \mathrm{C}\), the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and \(11.4\) torr, respectively.
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