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A 0.423 -g sample of an unknown gas exerts a pressure of 0.965 atm in a 1.00 -L container at \(445.7 \mathrm{~K}\). Calculate the molar mass of the gas.

Short Answer

Expert verified
The molar mass of the gas is approximately 16.02 g/mol.

Step by step solution

01

Write Down the Ideal Gas Law

The relationship between pressure, volume, temperature, and moles of a gas can be described using the ideal gas law, which is given by the equation: \[ PV = nRT \] where \( P \) is the pressure in atmospheres, \( V \) is the volume in liters, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant \( 0.0821 \, \text{L atm/mol K} \), and \( T \) is the temperature in Kelvin.
02

Convert Known Values to Proper Units

Make sure all values are in the correct units for the ideal gas law. Here, pressure \( P = 0.965 \, \text{atm} \), volume \( V = 1.00 \, \text{L} \), and temperature \( T = 445.7 \, \text{K} \). These values are already in the correct units.
03

Calculate Moles of Gas Using Ideal Gas Law

Rearrange the ideal gas law to solve for \( n \):\[ n = \frac{PV}{RT} \]Substitute the given values:\[ n = \frac{0.965 \, \text{atm} \times 1.00 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 445.7 \, \text{K}} \]Calculate the value:\[ n \approx 0.0264 \, \text{mol} \]
04

Calculate the Molar Mass of the Gas

Molar mass (\( M \)) can be calculated using the formula:\[ M = \frac{\text{mass of gas}}{n} \]where the mass of the gas is 0.423 grams and \( n = 0.0264 \, \text{mol} \). Substitute these values:\[ M = \frac{0.423 \, \text{g}}{0.0264 \, \text{mol}} \]Calculate the value:\[ M \approx 16.02 \, \text{g/mol} \]

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Key Concepts

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

Molar Mass Calculation
When you hear the term 'molar mass,' it simply means the mass of one mole of a substance. This is expressed in grams per mole (g/mol). To calculate the molar mass from an experiment, you need to know the mass of the sample and the amount of substance in moles. The formula used here is:
  • Molar mass (M) = mass of gas (in grams) / moles of gas (n)
In our example, we had a 0.423 g sample of gas. Using the ideal gas law, we calculated the moles of gas to be approximately 0.0264 mol. We then substituted these values into our molar mass formula:\[ M = \frac{0.423 \, \text{g}}{0.0264 \, \text{mol}} \approx 16.02 \, \text{g/mol} \]This tells us that each mole of the unknown gas has a mass of around 16.02 g.
Gas Law Equations
The Ideal Gas Law is a cornerstone of chemistry, allowing us to connect the macroscopic properties of gases. It's expressed as:\[ PV = nRT \]Here's what each symbol stands for:
  • P: Pressure of the gas, measured in atmospheres (atm).
  • V: Volume of the gas, measured in liters (L).
  • n: Amount of gas, measured in moles.
  • R: Ideal gas constant, approximately 0.0821 L atm/mol K.
  • T: Temperature of the gas, measured in Kelvin (K).
To solve problems using this law, you often need to rearrange it to find the unknown variable. For instance, to find the moles of gas, you would use:\[ n = \frac{PV}{RT} \]This manipulation is what allowed us to determine the number of moles in the original exercise, using values for pressure, volume, and temperature.
Conversion of Units
Before diving into calculations with gas laws, it’s essential to ensure all units are consistent and correct. Here’s a quick guide:
  • Pressure should be in atmospheres (atm). If you have pressure in other units like mmHg, you'll need to convert using 1 atm = 760 mmHg.
  • Volume should be in liters (L). For volume given in milliliters (mL), remember that 1 L = 1000 mL.
  • Temperature always uses Kelvin (K) for gas laws. Convert Celsius to Kelvin by adding 273.15.
Conversions are crucial since they ensure that you can directly apply the ideal gas constant ( R = 0.0821 L atm/mol K), which is key to solving these problems accurately. Luckily, in our exercise, the values were pre-converted properly, making it easy to input them directly into the equations.

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Most popular questions from this chapter

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4},\) can be made by the roomtemperature reaction of finely divided nickel metal with gaseous CO. This is the basis for purifying nickel on an industrial scale. If you have CO in a sealed \(1.50-\mathrm{L}\) flask at a pressure of \(418 \mathrm{mmHg}\) at \(25.0^{\circ} \mathrm{C},\) calculate the maximum mass in grams of \(\mathrm{Ni}(\mathrm{CO})_{4}\) that can be made.

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