Question

A \(1.20-\mathrm{g}\) sample of water is injected into an evacuated \(5.00-\mathrm{L}\) flask at \(65^{\circ} \mathrm{C}\). What percentage of the water will be vapor when the system reaches equilibrium? Assume ideal behavior of water vapor and that the volume of liquid water is negligible. The vapor pressure of water at \(65^{\circ} \mathrm{C}\) is \(187.5 \mathrm{mmHg}\).

Short answer

Approximately 66.37% of the water will be vapor at equilibrium.

Step-by-step solution

Calculate Moles of Water

First, determine the number of moles of water. Use the molar mass of water (about 18.015 g/mol) and the given mass of the water sample:\[ n = \frac{1.20 \, \text{g}}{18.015 \, \text{g/mol}} \approx 0.0666 \, \text{mol} \]

Convert Vapor Pressure to Atmospheres

The vapor pressure of water is given in mmHg. Convert this to atmospheres using the conversion factor (1 atm = 760 mmHg):\[ P_{\text{vap}} = \frac{187.5 \, \text{mmHg}}{760 \, \text{mmHg/atm}} \approx 0.246 \text{ atm} \]

Use Ideal Gas Law to Find Moles of Vapor

Apply the ideal gas law \( PV = nRT \) to find the moles of water vapor at equilibrium:\[ n_{\text{vapor}} = \frac{P_{\text{vap}} \times V}{R \times T} \]where \( V = 5.00 \, \text{L} \), \( R = 0.0821 \, \text{L atm/mol K} \), and \( T = 338 \text{ K} \) (since \( 65^{\circ} \text{C} = 338 \text{ K} \)). Therefore:\[ n_{\text{vapor}} = \frac{0.246 \, \text{atm} \times 5.00 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 338 \, \text{K}} \approx 0.0442 \, \text{mol} \]

Calculate Percentage of Water that is Vapor

Calculate the percentage of the initial 0.0666 mol of water that becomes vapor:\[ \text{Percentage of vapor} = \left( \frac{n_{\text{vapor}}}{n_{\text{initial}}} \right) \times 100 = \left( \frac{0.0442}{0.0666} \right) \times 100 \approx 66.37\% \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that expresses the relationship between pressure, volume, temperature, and the number of moles in a gaseous system. It is given by the formula:
\[ PV = nRT \]Here:

• \( P \) is the pressure of the gas in atmospheres (atm).
• \( V \) is the volume of the gas in liters (L).
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant, which equals \(0.0821 \, \text{L atm/mol K}\).
• \( T \) is the temperature in Kelvin (K).

In this context, we utilize the Ideal Gas Law to determine the amount of water vapor in the flask once it reaches equilibrium at the given conditions. The law assumes that water vapor behaves like an ideal gas, which simplifies calculations by allowing us to relate pressure and volume to the number of moles and temperature directly. This assumption is reasonable given that we are working with water vapor at relatively high temperatures.

Moles of Substance

Understanding moles is crucial when working with chemical equations and processes. Moles provide a way to quantify substance, allowing chemists to work with chemical reactions efficiently. A mole corresponds to Avogadro's number of particles, \(6.022 \times 10^{23}\), usually atoms or molecules.
In the given problem, you start by calculating the moles of water using its molar mass, which is about \(18.015 \, \text{g/mol}\). With a sample mass of \(1.20 \, \text{g}\), you can calculate the moles of water using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Through this calculation, you find \(0.0666 \, \text{mol}\) of water. This initial amount is essential for determining what portion of it undergoes vaporization under specified conditions.

Vaporization

Vaporization is the process whereby liquid water converts into vapor or gas. It requires energy input to overcome the intermolecular forces holding the water molecules together in liquid form, allowing them to escape into the gaseous phase.
In the exercise, the initial water sample is introduced into a flask, and as it heats up to \(65^{\circ} \text{C}\), some portion vaporizes. The vapor pressure of water at this temperature, given as \(187.5 \, \text{mmHg}\), indicates the pressure exerted by the water vapor when the liquid and vapor phases are in equilibrium.
Conversion to atmospheres helps integrate this value into the Ideal Gas Law. Understanding vaporization is key to calculating how much water transitions into the vapor phase, utilizing the vapor pressure and volume to find the extent of this transition.

Equilibrium

Equilibrium in chemistry occurs when the rate of forward reaction equals the rate of the reverse reaction, meaning no net change occurs in the concentrations of reactants and products. In the context of this exercise, equilibrium is reached between the liquid water and its vapor.
The system reaches equilibrium when the amount of water vapor in the flask reaches a pressure of \(0.246 \, \text{atm}\), matching the calculated vapor pressure. At this point, the rates of vaporization and condensation are equal.

• The system is dynamic, not static, meaning molecules continuously evaporate and condense.
• The percentage of the water that has vaporized is determined by comparing the moles of water vapor at equilibrium to the initial moles of water.

This understanding of equilibrium allows us to predict how much of the water sample remains as liquid and how much is in the vapor phase under the given conditions.