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Chapter 10: 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
The vapor pressure of the saltwater in the open beaker decreases over time. This is because evaporation removes water molecules from the liquid, leading to an increase in the salt concentration, which in turn lowers the vapor pressure due to fewer water molecules at the surface transitioning into the vapor phase.

Step by step solution

01

Understanding vapor pressure

Vapor pressure is the pressure of a vapor in contact with its liquid or solid form in a closed system. In this case, the vapor pressure is exerted by the water molecules in the presence of salt in the beaker.
02

Consider the salt in the water

Salt, a solute, dissolves in water, the solvent. This process affects the vapor pressure of the water by lowering it. This is because salt lowers the concentration of water molecules at the surface of the liquid, resulting in a lower number of water molecules transitioning into the vapor phase. This means that the vapor pressure caused by water molecules in the saltwater mixture is lower than that of pure water.
03

Evaporation and vapor pressure

As the saltwater beaker is left open in the room, evaporation occurs. During evaporation, the water molecules transition from the liquid phase to the vapor phase. The vapor pressure will change based on the concentration of water molecules in the liquid phase, which is affected by the presence of salt and the gradual loss of water molecules due to evaporation.
04

Open system vs. closed system

Since the beaker is considered an open system, the vapor above it is not in equilibrium with the surrounding atmosphere. This means that the vapor pressure of the saltwater system can change over time. In a closed system, the vapor pressure would eventually reach equilibrium, but this is not the case for the open saltwater beaker.
05

Effect of losing water molecules on vapor pressure

As the saltwater mixture loses water molecules over time due to evaporation, the concentration of salt in the remaining water in the beaker increases. This means that the vapor pressure of the remaining liquid phase decreases, as the water molecules are being replaced by salt particles at the surface. Therefore, the overall vapor pressure above the beaker will decrease over time. In conclusion, the vapor pressure of the saltwater in the open beaker decreases over time due to evaporation, which increases the concentration of salt in the water and lowers the concentration of water molecules at the surface. This process results in a lower vapor pressure above the beaker.

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Most popular questions from this chapter

Rubbing alcohol contains 585 g isopropanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) per liter (aqueous solution). Calculate the molarity.

How would you prepare 1.0 L of an aqueous solution of sodium chloride having an osmotic pressure of 15 atm at \(22^{\circ} \mathrm{C} ?\) Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

Rationalize the trend in water solubility for the following simple alcohols: $$\begin{array}{lc} \text { Alcohol } & \begin{array}{c} \text { Solubility } \\ \left(\mathrm{g} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { at } 20^{\circ} \mathrm{C}\right) \end{array} \\ \hline \text { Methanol, } \mathrm{CH}_{3} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Butanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{2} \mathrm{OH} & 8.14 \\ \text { Pentanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH} & 2.64 \\ \text { Hexanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH} & 0.59 \\ \text { Heptanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CH}_{2} \mathrm{OH} & 0.09 \\ \hline \end{array}$$

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?

For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)
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