Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0\) L of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

Short answer

The molar mass of the unknown gas is approximately 2.98 g/mol.

Step-by-step solution

Understand Graham's law of effusion

Graham's law of effusion states that the rate of effusion of two different gases is proportional to the square root of the inverse of their molar masses. Mathematically, this can be represented as: \( \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \) where \( Rate_1\) and \( Rate_2\) are the effusion rates of the two gases, \( M_1\) is the molar mass of the first gas, and \( M_2\) is the molar mass of the second gas. Since the rate of effusion and time required for effusion are inversely proportional, we can replace the rates in the equation with their corresponding times: \( \frac{Time_2}{Time_1} = \sqrt{\frac{M_2}{M_1}} \)

Set up the equation with the given information

Given the information about the effusion times of the unknown gas and the oxygen gas, set up the equation as follows: \( \frac{Time_{unknown}}{Time_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{unknown}}} \) Now plug in the given values: \( Time_{unknown} = 105 s \),\( Time_{O_2} = 31 s \) and \( M_{O_2} = 32 g/mol\), where \( M_{unknown}\) represents the molar mass of the unknown gas: \( \frac{105}{31} = \sqrt{\frac{32}{M_{unknown}}} \)

Solve for the molar mass of the unknown gas

To solve for the molar mass of the unknown gas, first square both sides of the equation: \( \left(\frac{105}{31}\right)^2 = \frac{32}{M_{unknown}} \) Now, solve for \( M_{unknown}\): \( M_{unknown} = \frac{32}{\left(\frac{105}{31}\right)^2} \) Calculate the molar mass: \( M_{unknown} ≈ 2.98 g/mol \) The molar mass of the unknown gas is approximately 2.98 g/mol.

Key concepts

Molar Mass Calculation

Understanding how to calculate molar mass is essential for solving problems related to the effusion of gases. Molar mass, symbolized as 'M', is the mass of one mole of a substance and its unit is grams per mole (g/mol). For elements, the molar mass is given on the periodic table as the atomic weight, and for compounds, it is the sum of the atomic weights of all atoms in the molecule.

In problems involving Graham's law of effusion, comparing the molar masses of two different gases can determine the rate at which they will effuse through a tiny opening. Typically, a known gas's molar mass is utilized to find the unknown gas's molar mass through a derived formula that relates effusion times and molar masses. As demonstrated in the exercise, recognizing the relationship between molar mass and effusion time is crucial for finding the molar mass of an unknown gas.

Effusion Rates

Effusion rate is a measure of how quickly a gas escapes through a small hole into a vacuum. It's an important concept in gas behavior, and in the context of Graham's law, it helps compare different gases' effusion rates. According to this law, lighter gases (with lower molar masses) effuse faster than heavier gases.

The rate is inversely proportional to the time taken for a specific volume of gas to effuse, meaning that the longer the effusion time, the slower the effusion rate. To determine effusion rates comparatively, it's often unnecessary to find the exact rate; understanding the ratio of the effusion rates, or times, can lead to solving for unknown variables such as the molar mass in the given exercise.

Gas Laws

Gas laws describe the behavior of gases and their interactions with temperature, pressure, volume, and amount. The laws governing the behavior of ideal gases are key to understanding gas effusion and diffusion processes. For instance, Boyle's Law posits an inverse relationship between pressure and volume, while Charles’s Law shows the direct proportionality between volume and temperature. Avogadro’s Law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

Graham's law integrates these concepts by assuming ideal gas behavior under constant pressure and temperature, which allows for the relationships. This simplification is what enables us to make direct comparisons between the effusion rates of two gases based on their molar masses, as seen in the exercise. However, it's important to note that real gases can deviate from ideal behavior under certain conditions, such as high pressures or low temperatures.