(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) . (b) Calculate the molar mass of a vapor that has a density of 7.135 \(\mathrm{g} / \mathrm{L}\) at \(12^{\circ} \mathrm{C}\) and 743 torr.

Short Answer

Expert verified
The density of sulfur hexafluoride gas at \(707 \, \mathrm{torr}\) and \(21^{\circ} \mathrm{C}\) is 6.75 g/L. The molar mass of the vapor with a density of 7.135 g/L at \(12^{\circ} \mathrm{C}\) and 743 torr is 215.39 g/mol.

Step by step solution

01

List the Known Variables

For this problem, we are given the pressure, P = 707 torr, the temperature, T = 21°C, and the molecule Sulfur Hexafluoride, SF6. We need to convert the given temperature to the kelvin scale. T(K) = T(°C) + 273.15
02

Convert Temperature to Kelvin

To convert the provided Celsius temperature to Kelvin: T(K) = 21 + 273.15 = 294.15 K
03

Convert Pressure from torr to atm

To use the ideal gas law equation, we need to convert the pressure from torr to atmospheres (atm). The conversion factor: 1 atm = 760 torr P(atm) = 707 torr / 760 torr = 0.9305 atm
04

Calculate Molar Mass of SF6

To find the molar mass of SF6, we need to add the atomic masses for one sulfur and six fluorine atoms: Molar Mass = (1 × 32.07 g/mol) + (6 × 19.00 g/mol) = 32.07 + 114 = 146.07 g/mol
05

Use Ideal Gas Law to Find the Moles

Now we can use the ideal gas law: PV = nRT 0.9305 atm * V = n * 0.0821 (L*atm/mol*K) * 294.15 K n = (0.9305 atm * V) / (0.0821 * 294.15 K)
06

Find Density

Density = (mass of the gas) / volume = (moles * molar mass) / volume Density = (n * 146.07 g/mol) / V Since n/V can be written as PV/RT, we can substitute it into the density equation. Density = ((0.9305 atm * 146.07 g/mol) / (0.0821 * 294.15 K))
07

Calculate Final Density

Calculate the final density: Density = (0.9305 * 146.07) / (0.0821 * 294.15) = 6.75 g/L So, the density of sulfur hexafluoride gas at \(707 \, \mathrm{torr}\) and \(21^{\circ} \mathrm{C}\) is 6.75 g/L. #Part (b): Calculating Molar Mass of Vapor#
08

List the Known Variables

In this part, we are given the density, \(\rho = 7.135 \, \mathrm{g/L}\), temperature, T = \(12^{\circ} \mathrm{C}\), and the pressure P = 743 torr. We want to find the molar mass of this vapor.
09

Convert Temperature to Kelvin

To convert the provided Celsius temperature to Kelvin: T(K) = 12 + 273.15 = 285.15 K
10

Convert Pressure from torr to atm

To use the ideal gas law equation, we need to convert the pressure from Torr to atmospheres (atm). The conversion factor: 1 atm = 760 torr P(atm) = 743 torr / 760 = 0.9776 atm
11

Rearranging Ideal Gas Law

We can rearrange the ideal gas law to solve for the molar mass (M). M = (Density * R * T) / P
12

Calculate Molar Mass

Calculate the molar mass of the gas: M = (7.135 g/L * 0.0821 L*atm/mol*K * 285.15 K) / 0.9776 atm = 215.39 g/mol Thus, the molar mass of the vapor is 215.39 g/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Density Calculation
Understanding how to calculate the density of a gas involves applying the ideal gas law, which relates the volume (V), pressure (P), temperature (T), and number of moles (n) of a gas. The formula for the ideal gas law is \( PV = nRT \), where R is the universal gas constant.

To find the density of a gas, one must start by finding the number of moles of the gas present in a certain volume. This number can be expressed by deriving \( n \) from the ideal gas law equation, with the relationship \( n = \frac{PV}{RT} \). Density (\( \rho \)) is mass per unit volume, so the equation becomes \( \rho = \frac{m}{V} \), where mass (m) is the product of the number of moles and molar mass (M). So, the density of a gas can be calculated using the formula \( \rho = \frac{nM}{V} \). By substituting the expression for \( n \) from the ideal gas law, the equation for density becomes \( \rho = \frac{P \times M}{RT} \).

For the step by step solution provided to calculate the density of sulfur hexafluoride gas, the molar mass of the gas is first determined, and then the ideal gas law is used to find the moles. Knowing the conditions under which the gas is present (temperature, volume, and pressure), the calculation of density then simplifies to substituting values into the rearranged ideal gas law.
Molar Mass Determination
Determining the molar mass of a substance is essential in many chemical calculations. Molar mass is the mass of one mole of a substance, and it is expressed in grams per mole (g/mol). In gases, the molar mass can be determined from the ideal gas law by rearranging the equation to solve for the molar mass (M). Given the density of a gas, the molar mass can be calculated using the formula \( M = \frac{\rho RT}{P} \).

This rearranged form exposes M as a function of the gas's density (\( \rho \)), the universal gas constant (R), the absolute temperature (T), and the pressure (P). This method is pivotal when analyzing vapors or gases for which the molar mass is not initially known. It is particularly useful in situations like the one demonstrated in the b part of the exercise, where the molar mass of a vapor is calculated from its density under specific temperature and pressure conditions. In this manner, knowing the properties of the gas enables us to back out important informational details about the substance's molar mass, therefore, facilitating further analysis or calculations.
Converting Temperature to Kelvin
The Kelvin scale is the base unit of temperature in the International System of Units (SI) and is a crucial component in gas calculations, particularly when using the ideal gas law. Unlike Celsius or Fahrenheit, Kelvin is an absolute temperature scale where zero Kelvin (0 K) is absolute zero, the theoretical point where particles have minimal thermal motion.

To convert Celsius to Kelvin, the equation \( T(K) = T(°C) + 273.15 \) is used. This step is vital as all gas law equations require temperature to be expressed in Kelvin to ensure consistency and accuracy. For example, in the solution provided, temperatures given in Celsius are converted to Kelvin before being plugged into the ideal gas law. This conversion ensures that temperature is viewed on an absolute scale, allowing direct proportionality between variables like pressure, volume, and the number of moles in the gas law, which would not be the case with Celsius or Fahrenheit scales.

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Most popular questions from this chapter

The temperature of a \(5.00-\mathrm{L}\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\) . If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean- square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls per second.

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\text { mfp }}\) , like the ideal-gas constant) and define units for \(R_{\operatorname{mfp}}\) .

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? \begin{equation}\begin{array}{l}{\text { (a) Because Xe is a noble gas, there is less tendency for the Xe }} \\ {\text { atoms to repel one another, so they pack more densely in }} \\ {\text { the gaseous state. }} \\ {\text { (b) Xe atoms have a higher mass than } \mathrm{N}_{2} \text { molecules. Because }} \\ {\text { both gases at STP have the same number of molecules per }} \\ {\text { unit volume, the Xe gas must be denser. }}\\\\{\text { (c) The Xe atoms are larger than } \mathrm{N}_{2} \text { molecules and thus take }} \\ {\text { up a larger fraction of the space occupied by the gas. }} \\\ {\text { (d) Because the Xe atoms are much more massive than the }} \\\ {\mathrm{N}_{2} \text { molecules, they move more slowly and thus exert }} \\\ {\text { less upward force on the gas container and make the gas }} \\ {\text { appear denser. }}\end{array}\end{equation}

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