Problem 35 - General Chemistry: Principles and Modern Applications

Question

A 25.0 L volume of \(\mathrm{He}(\mathrm{g})\) at \(30.0^{\circ} \mathrm{C}\) is passed through \(6.220 \mathrm{g}\) of liquid aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) at \(30.0^{\circ} \mathrm{C} .\) The liquid remaining after the experiment weighs \(6.108 \mathrm{g}\) Assume that the He(g) becomes saturated with aniline vapor and that the total gas volume and temperature remain constant. What is the vapor pressure of aniline at \(30.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The vapor pressure of aniline at \(30.0^{\circ} C\) is \(0.012 atm\)

Step 1: Determine the amount of aniline that has vaporized

From the given weights, we can find the amount of aniline that has vaporized by subtracting the final weight of the remaining liquid aniline from the initial weight, which yields \(6.220g - 6.108g = 0.112g\) aniline vapor.

Step 2: Convert the weight of aniline vapor to moles

Next, we will convert this weight of aniline vapor into moles, using the molar mass of aniline which is about \(93.13g/mol.\) Hence, the number of moles of aniline vapor is \(0.112g ÷ 93.13g/mol = 1.202 x 10^{-3} moles\).

Step 3: Apply the Ideal Gas law

The ideal gas law states that P = nRT/V. Here, P is the vapor pressure (which we are looking for), n is the number of moles of aniline in the gaseous state, R is the gas constant (0.0821 Latm/mol.K for this problem), T is the temperature in Kelvin, and V is the volume. However, it is crucial to convert the given temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. Thus, T = 30°C + 273.15 = 303.15 K.

Step 4: Calculate the vapor pressure of aniline

Substitute the known values into the ideal gas law to solve for the vapor pressure P. P = (1.202 x 10^{-3} mol * 0.0821 L.atm/mol.K * 303.15K) / 25.0L. Solving for P gives \(P = 0.012 atm\)

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle used to relate the pressure, volume, temperature, and number of moles of a gas. This law is expressed with the formula: \[ P = \frac{nRT}{V} \]where:

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

This equation helps to calculate the vapor pressure of gases, which is essential in understanding how gases behave under different conditions.
For example, if the temperature and volume are kept constant, changing the number of moles directly affects the pressure of the gas. For our problem, we had to rearrange this equation to find the vapor pressure \(P\) of the aniline vapor formed.

Moles Conversion

Converting grams to moles is a pivotal step in calculating any gas-related problem. To perform this conversion, you'll use the substance's molar mass. Molar mass is the weight of one mole of a given element or compound, usually expressed in \(\mathrm{g/mol}\).
The formula for conversion is:\[\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar Mass}} \]For our problem involving aniline, the given mass was \(0.112 \mathrm{g}\), and the molar mass of aniline is \(93.13 \mathrm{g/mol}\).
By applying the formula above:\[\frac{0.112 \mathrm{g}}{93.13 \mathrm{g/mol}} = 1.202 \times 10^{-3} \text{ moles of aniline} \]This conversion is critical. It allows us to find out how much of the aniline has transitioned into its gaseous state. Knowing the number of moles will enable us to calculate the pressure using the Ideal Gas Law.

Vaporization

Vaporization is the process where liquid transitions to a vapor. This occurs when molecules in a liquid gain enough energy to break away into the gaseous phase.
In our exercise, helium gas was used to facilitate the vaporization of aniline.The vaporization process involved calculating how much aniline transformed from liquid to vapor:

• Initial liquid aniline mass: \(6.220 \mathrm{g}\)
• Remaining liquid aniline mass: \(6.108 \mathrm{g}\)
• Mass of vaporized aniline: \(6.220 \mathrm{g} - 6.108 \mathrm{g} = 0.112 \mathrm{g}\)

This small loss in mass represents the amount of aniline that vaporized.Understanding vaporization is crucial for determining how much of a liquid has become a gas, especially when calculating related properties like vapor pressure.

Vapor Pressure Calculation

Vapor pressure indicates how much pressure a vapor exerts. It's a measure of the tendency of a liquid to evaporate. For gases, knowing the vapor pressure is essential for understanding how volatile a substance is at a specific temperature.
To calculate vapor pressure:

• Determine the number of moles of the vaporized substance.
• Apply the Ideal Gas Law formula, \( P = \frac{nRT}{V} \), shifting variables as needed.

In the example, we calculated the vapor pressure of aniline with given values:

• \(n = 1.202 \times 10^{-3} \text{ moles}\).
• \(R = 0.0821 \space \text{L.atm/mol.K}\).
• \(T = 303.15 \text{ K}\).
• \(V = 25.0 \text{ L}\).
• Substituting those into the formula to find \( P \), we get: \( P = \frac{1.202 \times 10^{-3} \phantom{a} \text{mol} \times 0.0821 \text{ L.atm/mol.K} \times 303.15 \text{ K}}{25.0 \text{ L}} = 0.012 \text{ atm}\).
• This result gives us a clear picture of the vapor pressure of aniline at \(30^{\circ} C\).