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(a) Calculate the density of sulfur hexafluoride gas at 94.26 \(\mathrm{kPa}\) 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 \(99.06 \mathrm{kPa}\)..

Short Answer

Expert verified
The density of sulfur hexafluoride gas at 94.26 kPa and 21°C is approximately 6.115 g/L. The molar mass of the vapor that has a density of 7.135 g/L at 12°C and 99.06 kPa is approximately 153.1 g/mol.

Step by step solution

01

Part (a) - Calculate the density of sulfur hexafluoride gas

To calculate the density of sulfur hexafluoride, we'll use the Ideal Gas Law formula: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the given pressure and temperature to the appropriate SI units: Pressure (P) = 94.26 kPa Temperature (T) = 21 °C + 273.15 = 294.15 K Additionally, we'll need to know the molar mass of sulfur hexafluoride (SF6). From the periodic table, the molar mass of sulfur is 32 g/mol, and the molar mass of fluorine is 19 g/mol. Therefore, the molar mass of sulfur hexafluoride is: Molar mass of SF6 = (\(1\times32\)) + (\(6\times19\)) = 32 + 114 = 146 g/mol Now we can use the Ideal Gas Law formula in terms of density (\(\rho\)): \(\rho = \frac{P\times M}{R\times T}\) Where \(\rho\) is the density, P is the pressure, M is the molar mass, R is the ideal gas constant, and T is the temperature in Kelvin. The value of R is 8.314 J/(mol K), but for this calculation, we'll use R = 0.0821 L atm / (mol K) (to match the pressure unit).
02

Part (a) - Calculate the density

Now we can fill in the values: \(\rho = \frac{(94.26\,\text{kPa}) \frac{101.3\,\text{Pa}}{1\,\text{kPa}} \times \frac{1\, \mathrm{atm}}{101325\,\text{Pa}} \times (146\,\text{g/mol})}{(0.0821\,\text{L atm} /(\text{mol K})) \times (294.15\, \text{K})}\) Calculate the density: \(\rho \approx 6.115\, \text{g/L}\) The density of sulfur hexafluoride gas at 94.26 kPa and 21°C is approximately 6.115 g/L.
03

Part (b) - Calculate the molar mass of a vapor

As given, we'll need to calculate the molar mass of a vapor with: Density (\(\rho\)) = 7.135 g/L Temperature (T) = 12 °C + 273.15 = 285.15 K Pressure (P) = 99.06 kPa We can rearrange the Ideal Gas Law formula in terms of molar mass (M): \(M = \frac{\rho \times R \times T}{P}\)
04

Part (b) - Calculate the molar mass

Now we can fill in the values: \(M = \frac{(7.135\, \text{g/L}) \times (0.0821\, \text{L atm} /(\text{mol K})) \times (285.15\, \text{K})}{(99.06\,\text{kPa}) \frac{101.3\,\text{Pa}}{1\,\text{kPa}} \times \frac{1\, \mathrm{atm}}{101325\,\text{Pa}}}\) Calculate the molar mass: \(M \approx 153.1\, \text{g/mol}\) The molar mass of the vapor that has a density of 7.135 g/L at 12°C and 99.06 kPa is approximately 153.1 g/mol.

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Key Concepts

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

Density Calculation
Understanding how to calculate the density of a gas using the Ideal Gas Law is an essential skill in chemistry. The density (\(\rho\)) of a gas is a measure of how much mass is present in a given volume. The Ideal Gas Law, stated as\[ PV = nRT \]can be rearranged to calculate density in the following form: \[ \rho = \frac{P \times M}{R \times T} \]Here:
  • **\(P\)** is the pressure of the gas.
  • **\(M\)** is the molar mass.
  • **\(R\)** is the universal gas constant.
  • **\(T\)** is the temperature in Kelvin.
To perform this calculation accurately, make sure to convert all your units to match the Ideal Gas Law's requirements. For example, pressure should be in atm, and temperature should be in Kelvin. When handling gases like sulfur hexafluoride, knowing its molar mass is crucial. The calculation involves substituting known values into the density formula to find the gas's density under specified conditions. This concept is not only crucial in academic exercises but also in real-world applications where gases are used.
Molar Mass Determination
Determining the molar mass of a vapor requires a good understanding of gas behavior and the Ideal Gas Law. The molar mass (\(M\)) is an important property of a substance and is used to determine how heavy a molecule is. This property is calculated using the rearranged Ideal Gas Law:\[ M = \frac{\rho \times R \times T}{P} \]Where:
  • **\(\rho\)** is the density of the gas.
  • **\(R\)** is the ideal gas constant.
  • **\(T\)** is the temperature in Kelvin.
  • **\(P\)** is the pressure of the gas.
To find the molar mass, you insert the given values into this equation. Be meticulous with unit conversions – converting temperature to Kelvin and pressure to atm if necessary. Understanding how to calculate molar mass is crucial for identifying an unknown gas or confirming the purity of a sample. These calculations are often used in chemical industries and laboratory settings to ensure processes are running with the correct materials.
Gas Behavior Under Varying Conditions
Gases behave differently under varying temperature, pressure, and volume conditions. The Ideal Gas Law helps to predict and explain these behaviors. By manipulating the equation \[ PV = nRT \]gases can be studied in numerous conditions by keeping some variables constant, such as in Boyle’s Law, Charles's Law, and Gay-Lussac's Law. - **As temperature increases**, if pressure is constant, gas volume increases.- **With increasing pressure**, if temperature is constant, gas volume decreases.These relationships help understand concepts such as:- Why hot air balloons rise as heating the air inside increases its volume, decreasing density.- High altitude, where pressure is lower and thus lower density, affects how quickly gases expand or contract.Grasping these varying behaviors of gases is crucial to predicting reactions, adjusting industrial processes, and designing systems where gases' properties are utilized. Mastering the Ideal Gas Law helps in controlling and utilizing gas behaviors practically and is part of the broader study of thermodynamics.

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Most popular questions from this chapter

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