Question

Carbon dioxide, \(\mathrm{CO}_{2},\) was shown to effuse through a porous plate at the rate of \(0.033 \mathrm{mol} / \mathrm{min.}\) The same quantity of an unknown gas, 0.033 moles, is found to effuse through the same porous barrier in 104 seconds. Calculate the molar mass of the unknown gas.

Short answer

The molar mass of the unknown gas is approximately 132.00 g/mol.

Step-by-step solution

Understand Graham's Law of Effusion

Graham's Law of Effusion describes the relation between the rates of effusion for two gases and their molar masses. It states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as: \[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]where \(\text{Rate}_1\) and \(\text{Rate}_2\) are the effusion rates and \(M_1\) and \(M_2\) are the molar masses of the two gases.

Calculate Rates of Effusion

First, calculate the effusion rates for both gases. The rate of effusion for carbon dioxide, \(\text{Rate}_{\mathrm{CO}_2}\), is given as \(0.033 \text{ mol/min}\). For the unknown gas, convert the time from seconds to minutes: \[ 104 \text{ s} = \frac{104}{60} \text{ min} = 1.733 \text{ min} \].Then, calculate the rate of effusion for the unknown gas via \[ \text{Rate}_{\text{unknown}} = \frac{0.033 \, \text{mol}}{1.733 \, \text{min}} \approx 0.01905 \, \text{mol/min} \].

Apply Graham's Law to Find Molar Mass

Using Graham's Law, substitute both effusion rates and solve for the molar mass \( M \) of the unknown gas:\[ \frac{\text{Rate}_{\mathrm{CO}_2}}{\text{Rate}_{\text{unknown}}} = \sqrt{\frac{M_{\text{unknown}}}{M_{\mathrm{CO}_2}}} \]Substitute the known values and the molar mass of CO2, \( M_{\mathrm{CO}_2} = 44.01 \, \text{g/mol}\):\[ \frac{0.033}{0.01905} = \sqrt{\frac{M_{\text{unknown}}}{44.01}} \]Calculate the left side approximately:\[ 1.732 \approx \sqrt{\frac{M_{\text{unknown}}}{44.01}} \].

Solve for Unknown Molar Mass

Square both sides of the equation from Step 3 to eliminate the square root:\[ 1.732^2 = \frac{M_{\text{unknown}}}{44.01} \].Calculate:\[ 2.999 = \frac{M_{\text{unknown}}}{44.01} \].Finally, solve for \( M_{\text{unknown}} \):\[ M_{\text{unknown}} = 2.999 \times 44.01 \approx 132.00 \, \text{g/mol} \].

Key concepts

Effusion Rate Calculation

Effusion rate calculation is a fundamental concept within Graham's Law of Effusion. It refers to determining the rate at which gas particles pass through a tiny opening or porous barrier. To understand this concept, you need to consider how quickly each gas effuses compared to another. The rate is often given in moles per minute or another time unit. For example, in our exercise, the effusion rate of carbon dioxide (\(\text{Rate}_{\mathrm{CO}_2}\)) is known to be 0.033 mol/min. When the time taken for an unknown gas to pass through the same porous plate is considered, we convert it from 104 seconds to minutes, resulting in approximately 1.733 minutes. Subsequently, the effusion rate of the unknown gas is calculated using this time period.Calculating effusion rates involves dividing the number of moles of gas by the time taken for those moles to effuse:

• Rate of known gas: 0.033 mol/min
• Rate of unknown gas: 0.033 mol / 1.733 min ≈ 0.01905 mol/min

This allows us to directly apply Graham's Law of Effusion to compare their behaviors.

Molar Mass Determination

Understanding molar mass determination through Graham's Law involves comparing effusion rates of two gases. This method provides an indirect, yet effective, way to calculate the molar mass of an unknown gas when the effusion rate is known. The mathematical formula used is:\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]Here, \(\text{Rate}_1\) represents the effusion rate of the known gas, while \(\text{Rate}_2\) is the effusion rate for the unknown gas. \(M_1\) and \(M_2\) are the molar masses of the known and unknown gases, respectively. In our example, using the rates of CO2 and the unknown gas, we can rearrange Graham's Law to solve for the unknown molar mass as follows:Begin with:\[ \frac{0.033}{0.01905} = \sqrt{\frac{M_{\text{unknown}}}{44.01}} \]By squaring each side of the equation, you eliminate the square root and make it possible to solve for the unknown molar mass, hence:\[ 1.732^2 = \frac{M_{\text{unknown}}}{44.01} \]When squared, solve: ul>

Finds \(M_{\text{unknown}} = 2.999 \cdot 44.01 \approx 132.00 \text{ g/mol}\)

Effusion method allows such calculations when direct measurement is not an option.

Porous Plate Effusion

Porous plate effusion is a practical way to study the movement of gases by observing how they pass through a material with tiny holes. It is crucial in demonstrating Graham's Law of Effusion in experiments. This method relies on the natural tendency of gases to travel from a region of higher concentration to a lower one through tiny openings. In the context of porous plates, effusion can be monitored by measuring the time taken for a specific amount of gas to move through the barrier. We see this in exercises where different gases show variation in effusion rates. The porous plates ensure that effusion is the sole process occurring, minimizing other factors like diffusion.For educational examples, understanding gases like carbon dioxide (\(\mathrm{CO}_2\)) simplifies explanations since its molar mass and typical behaviors are well-documented. Comparing its effusion with an unknown gas using a porous plate allows insightful conclusions about molecular characteristics. Using such methods permits deeper comprehension not just of gas behaviors but also enhances grasp of fundamental gas laws like Graham's, assisting in diverse settings from laboratories to industrial applications.