Question

If \(4.83 \mathrm{~mL}\) of an unknown gas effuses through a hole in a plate in the same time it takes \(9.23 \mathrm{~mL}\) of argon, \(\mathrm{Ar}\), to effuse through the same hole under the same conditions, what is the molecular weight of the unknown gas?

Short answer

The molecular weight of the unknown gas is approximately 145.96 g/mol.

Step-by-step solution

Understand Effusion and Graham's Law

Effusion is the process by which gas molecules pass through a tiny hole into a vacuum. Graham's Law states that the ratio of the rates of effusion of two gases is equal to the inverse square root of the ratio of their molar masses. Mathematically, it's expressed as: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) and \( r_2 \) are the effusion rates of gases 1 and 2, and \( M_1 \) and \( M_2 \) are their molar masses.

Relate Rates of Effusion to Volumes

The rate of effusion of a gas is proportional to the volume of gas that effuses in a given time. Given that 4.83 mL of the unknown gas effuses in the same time as 9.23 mL of argon, the ratio \( \frac{r_{\text{unknown}}}{r_{\text{Ar}}} = \frac{4.83}{9.23} \) can be determined.

Apply Graham's Law

Using Graham's Law \( \frac{r_{\text{unknown}}}{r_{\text{Ar}}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{unknown}}}} \), substitute \( \frac{4.83}{9.23} \) for the ratio of effusion rates, and \( M_{\text{Ar}} = 39.95 \) g/mol for the molar mass of argon.

Calculate the Molecular Weight

Rearrange the formula to solve for the unknown molar mass: \( M_{\text{unknown}} = M_{\text{Ar}} \left( \frac{9.23}{4.83} \right)^2 \). Calculate the result to find the molar mass of the unknown gas.

Compute the Answer

First, compute the ratio: \( \frac{9.23}{4.83} \approx 1.9114 \). Then, square the ratio: \( (1.9114)^2 \approx 3.6534 \). Finally, multiply by the molar mass of argon: \( 39.95 \times 3.6534 \approx 145.96 \). The molecular weight of the unknown gas is approximately 145.96 g/mol.

Key concepts

Graham's Law

Graham's Law provides a way to compare the effusion rates of different gases by relating these rates to their molar masses. Effusion is the process where gas particles pass through a small opening. The law can be expressed with the formula: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \). Here, \( r_1 \) and \( r_2 \) are the effusion rates of two gases, and \( M_1 \) and \( M_2 \) are their molar masses.
This inverse relationship means that the lighter a gas is, the faster it will effuse compared to a heavier gas. If you know the effusion rates and one molar mass, you can solve for the unknown molar mass. Understanding this principle helps compare how quickly different gases escape through a barrier.

Molecular Weight Calculation

To find the molecular weight of a gas, you often rely on a known reference, like in our problem, where argon is the reference gas. Start by applying Graham's Law and setting up the ratio of effusion rates. You get \( \frac{r_{\text{unknown}}}{r_{\text{Ar}}} = \frac{4.83}{9.23} \). This equation gives the comparative rate of effusion.
To solve for the unknown molar mass, rearrange Graham's Law: \( M_{\text{unknown}} = M_{\text{Ar}} \left( \frac{r_{\text{Ar}}}{r_{\text{unknown}}} \right)^2 \). Substitute the known values: \( M_{\text{Ar}} = 39.95 \text{ g/mol}\) and the rate ratio. Calculating through yields the molecular weight you want to find.

Effusion Rate

The effusion rate is determined by how much gas passes through a tiny hole over a given time period. In this context, its ratio is linked directly to the volume of gas that escapes. For our gases, the effusion rate can be expressed as the ratio of the volumes: \( \frac{4.83}{9.23} \). Here, the gas that exits slower has the larger molar mass.
Using this concept, it’s clear that effusion rates are crucial to compare unknown gases, revealing more about their characteristics when paired with Graham's Law. Understanding the concept of effusion rates helps in determining how gases behave under similar conditions.

Molar Mass of Gases

The molar mass of gases is a measure of how much a mole of gas weighs, providing insight into the gas's molecular weight and structure. It is essential in converting between grams and moles, and it plays a critical role when applying Graham's Law of effusion.
In the exercise provided, the molar mass of argon was used as a known reference (39.95 g/mol). By applying Graham's Law, we derived the unknown gas’s molar mass based on its effusion behavior. Determining the molar mass helps understand a gas's density and its chemical properties relative to other gases.