An unknown gas \(X\) effuses 1.65 times faster than \(\mathrm{C}_{3} \mathrm{H}_{8}\). What is the molecular mass of gas \(X ?\)

The concept of rate of effusion is a fundamental aspect of gas behavior described by Graham's Law. It refers to how quickly a gas escapes through a tiny hole or porous barrier into a vacuum or less pressurized space. Imagine you have a balloon with a small puncture; the speed at which the air streams out of this puncture is analogous to the rate of effusion.

To compare two different gases, Graham's Law provides a way to understand their effusion rates relative to each other. According to this law, lighter gases (those with a lower molar mass) effuse more rapidly than heavier gases. This is because lighter gas molecules travel faster on average, making them more likely to collide with and pass through small openings.

When it comes to finding the molecular mass (often referred to as molar mass when calculated per mole of substance), it's about totaling the atomic masses of all the atoms in a molecule. Taking propane, \( \mathrm{C}_{3}\mathrm{H}_{8} \), as an example, we add three times the atomic mass of carbon to eight times the atomic mass of hydrogen to determine its molecular mass.

Here's how you might visualize the process:

• Atomic mass of Carbon (C): 12.01 g/mol
• Atomic mass of Hydrogen (H): 1.008 g/mol
• Molecular mass of \( \mathrm{C}_{3}\mathrm{H}_{8} \): \( 3 \times 12.01 + 8 \times 1.008 = 44.095 \) g/mol

It's this calculated value that plays a crucial part in applying Graham's Law for effusion rates and subsequently identifying unknown gases.

The gas effusion rate is an expression of how quickly a gas exits an enclosed space. In Graham's Law, this rate is directly tied to the molar mass of the gas. If you know the effusion rate of one gas and the molar masses of both gases involved, you can find the effusion rate of the second gas.

For instance, if an unknown gas effuses 1.65 times faster than propane, you just need the molar mass of propane to calculate the unknown gas's molar mass using the formula from Graham's Law. Thus, by manipulating the relevant formula, you can derive the gas effusion rate in relation to another known substance.

The molar mass of gas directly influences effusion rates according to Graham's Law. This is because the kinetic theory of gases suggests that all gases at a given temperature have the same kinetic energy on average. As a consequence, a lower molar mass means higher velocity, leading to a greater rate of effusion.

To put it in perspective, let's consider the example where an unknown gas effuses 1.65 times faster than propane. Knowing propane's molar mass allows us to deduce the molar mass of this unknown gas. Following the rearranged equation from Graham's Law, you can solve for the unknown gas's molar mass, \( M_X = \frac{44.095}{(1.65)^2} = 16.18 \text{g/mol} \). This calculation shows a direct link between a gas's molar mass and its effusion behavior which is crucial for identifying gases and understanding their physical properties.