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At 741 torr and \(44^{\circ} \mathrm{C}, 7.10 \mathrm{~g}\) of a gas occupies a volume of \(5.40 \mathrm{~L}\). What is the molar mass of the gas?

Short Answer

Expert verified
The molar mass of the gas is approximately 35.86 g/mol.

Step by step solution

01

Convert Units

Ensure that all units are compatible with the Ideal Gas Law. Convert the temperature to Kelvin by adding 273.15 to the Celsius temperature: \( T = 44 + 273.15 = 317.15 \ K \). Similarly, convert the pressure from torr to atm by dividing by 760: \( P = \frac{741}{760} = 0.975 \ \text{atm} \).
02

Apply the Ideal Gas Law

The Ideal Gas Law is given by \( PV = nRT \). We know the pressure \( P = 0.975 \ \text{atm} \), volume \( V = 5.40 \ \text{L} \), temperature \( T = 317.15 \ K \), and the ideal gas constant \( R = 0.0821 \ \text{L atm/mol K} \).
03

Solve for Moles \( n \)

Rearrange the Ideal Gas Law to solve for \( n \): \( n = \frac{PV}{RT} \). Substitute the known values: \( n = \frac{0.975 \times 5.40}{0.0821 \times 317.15} \approx 0.198 \ \text{moles} \).
04

Calculate Molar Mass

The molar mass \( M \) is given by \( M = \frac{\text{mass of the gas}}{n} \). Given that the mass is \( 7.10 \ \text{g} \) and \( n = 0.198 \ \text{moles} \), we find \( M = \frac{7.10}{0.198} \approx 35.86 \ \text{g/mol} \).

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Key Concepts

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

Molar Mass Calculation
The molar mass of a gas is a measure of how much one mole of that substance weighs. Understanding how to calculate the molar mass is essential for chemists and students because it helps in identifying an unknown gas.
In the exercise provided, you calculated the molar mass by first finding the number of moles of the gas. The number of moles \( n \) was found using the Ideal Gas Law, which relates pressure, volume, and temperature to moles. The equation used is:
  • \( n = \frac{PV}{RT} \)
Here, \( P \) is the pressure, \( V \) is the volume, \( T \) is the temperature of the gas, and \( R \) is the ideal gas constant.
Once you have \( n \), the molar mass \( M \) can be calculated by dividing the mass of the gas by the moles of the gas:
  • \( M = \frac{\text{mass of the gas}}{n} \)
This step reveals how much one mole of the unknown gas weighs, providing you with its molar mass.
Gas Laws
Gas laws are mathematical relationships governing the behavior of gases. Among the most crucial of these is the Ideal Gas Law, which was used in the exercise. The Ideal Gas Law is expressed as \( PV = nRT \).
This law essentially combines several other gas laws into one comprehensive formula: Boyle's Law, Charles's Law, and Avogadro's Law. Each of these describes the relationship between pairs of physical properties of a gas:
  • Boyle's Law: Pressure and volume are inversely proportional at constant temperature.
  • Charles's Law: Volume and temperature are directly proportional at constant pressure.
  • Avogadro's Law: Volume is directly proportional to the number of moles of gas at constant temperature and pressure.
By using these concepts together in the Ideal Gas Law, it's possible to calculate unknown values of a gas' state, like the number of moles or volume, given other properties.
This is particularly useful in determining unknown characteristics of gases in experimental setups, like calculating the molar mass.
Unit Conversion
Unit conversion is an essential skill in chemistry, especially when applying formulas like the Ideal Gas Law. The exercise required converting different units to match those in the gas law constant \( R \), which is \( 0.0821 \ \text{L atm/mol K} \).
First, we needed to convert temperature from Celsius to Kelvin. This is necessary because Kelvin is the unit used in scientific calculations and the Ideal Gas Law. To convert Celsius to Kelvin, simply add 273.15 to the Celsius value:
  • \( T(K) = T(°C) + 273.15 \)
Pressure often needs conversion, such as from torr to atm, as seen in the exercise. The conversion is achieved by dividing the pressure in torr by 760, given that 1 atm is equivalent to 760 torr:
  • \( P(atm) = \frac{P(torr)}{760} \)
Converting units ensures the results will be consistent and comprehensible in scientific measurements.
This is crucial for accurate calculations and helps prevent common errors that occur when different unit systems are mixed.

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