/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 A gas of unknown molecular mass ... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The molar mass of the unknown gas is approximately 9.96 g/mol.

Step by step solution

01

Write the formula for Graham's Law of Effusion in terms of time and molar mass

Given that rate and time are inversely proportional, we can rewrite Graham's Law of Effusion as the ratio of the square root of the molar masses instead of the effusion rates. \[ \frac{\sqrt{MM_{1}}}{\sqrt{MM_{2}}} = \frac{t_{2}}{t_{1}} \]Where: - \(MM_{1}\) is the molar mass of the unknown gas - \(MM_{2}\) is the molar mass of Oâ‚‚ gas - \(t_{1}\) is the time for the unknown gas to effuse (105 s) - \(t_{2}\) is the time for the Oâ‚‚ gas to effuse (31 s)
02

Insert the given values and rearrange the formula to solve for \(MM_{1}\)

We know the molar mass of Oâ‚‚ gas is 32 g/mol. Plugging in the given values into the formula, we get:\[ \frac{\sqrt{MM_{1}}}{\sqrt{32}} = \frac{31}{105} \]Now, we will rearrange the formula to solve for the molar mass of the unknown gas, \(MM_{1}\):\[ \sqrt{MM_{1}} = \frac{31}{105} \times \sqrt{32} \]Square both sides of the equation to get rid of the square root:\[ MM_{1} = \left(\frac{31}{105} \times \sqrt{32}\right)^2 \]
03

Calculate the molar mass of the unknown gas

Now, we can calculate the value of \(MM_{1}\) using the rearranged formula:\[ MM_{1} = \left(\frac{31}{105} \times \sqrt{32}\right)^2 \Rightarrow MM_{1} \approx 9.96 \] So, the molar mass of the unknown gas is approximately 9.96 g/mol.

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Key Concepts

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

Molar Mass
Molar mass is a critical concept in chemistry. It's defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Understanding molar mass can be immensely helpful when working with reactions and calculations involving gases.
Every element has its own specific molar mass. To determine the molar mass of a compound, you add up the molar masses of all the atoms in that compound.
When it comes to gases, knowing the molar mass is essential for performing various calculations, such as predicting reactions or calculating effusion rates. For example, in our exercise, we needed the molar mass of oxygen to solve for the unknown gas using Graham's Law.
  • Molar mass provides a link between a substance's physical mass and the amount of substance represented in moles.
  • It helps in converting between grams and moles, supporting stoichiometric calculations.
  • The periodic table can be used to find the molar mass of each element, which aids in calculating the molar mass of compounds.
Getting familiar with molar mass enables us to delve deeper into other gas-related calculations and laws, such as the understanding of effusion rates.
Effusion Rate
The effusion rate refers to the speed at which a gas escapes through a small opening, a concept closely examined through Graham's Law. Effusion is specifically the process when gas particles escape through the tiny gaps without collisions among gas molecules.
For gases, the rate of effusion depends significantly on the molar mass of the gas and operates under specific pressure conditions.
The main takeaway is that lighter gases effuse more rapidly than heavier gases. This understanding allows us to predict and compare how various gases will behave under similar conditions.
  • Effusion occurs when a gas passes through a small hole into a vacuum, without interacting with other particles.
  • The relationship between effusion rates and time is notably inversely proportional: faster effusion rates mean less time required for a given volume of the gas to effuse.
  • This concept forms the foundation for using Graham's Law for comparative calculations and solving unknowns in gas equations.
The effusion rate is an integral part of understanding how gases behave, providing experimental and theoretical insights into molecular motion and gas dynamics.
Graham's Law Formula
Graham's Law of Effusion is a fundamental principle in chemistry that relates the effusion rates of two gases to their molar masses. This law is particularly useful for comparing gases under identical conditions.
Graham's Law states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be expressed as:
\[\frac{r_1}{r_2} = \sqrt{\frac{MM_2}{MM_1}}\]
However, if we know the times taken for the gases to effuse, it can also be expressed in terms of time as:
\[\frac{\sqrt{MM_1}}{\sqrt{MM_2}} = \frac{t_2}{t_1}\]
In our specific exercise, we calculated the molar mass of an unknown gas by manipulating the formula based on the effusion times of the unknown gas and oxygen. This rearrangement helped solve for the unknown gas's molar mass effectively.
  • The inverse square root relationship with molar mass means heavier gases effuse slower than lighter ones.
  • Using this formula enables predicting unknowns or verifying the behavior of gases in experimental setups.
  • The formula is powerful because it links fundamental physical properties (molar mass) with observable experimental data (effusion rates or times).
Understanding and using Graham's Law gives crucial insight into blending theoretical chemistry concepts with practical experimental investigation.

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Most popular questions from this chapter

A set of bookshelves rests on a hard floor surface on four legs, each having a cross-sectional dimension of \(3.0 \times 4.1 \mathrm{~cm}\) in contact with the floor. The total mass of the shelves plus the books stacked on them is \(262 \mathrm{~kg} .\) Calculate the pressure in pascals exerted by the shelf footings on the surface.

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ At a certain temperature and pressure, \(1.2 \mathrm{~L}\) of \(\mathrm{N}_{2}\) reacts with \(3.6 \mathrm{~L}\) of \(\mathrm{H}_{2}\). If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3}\), at the same temperature and pressure, will be produced?

Perform the following conversions: (a) 0.912 atm to torr, (b) 0.685 bar to kilopascals, (c) \(655 \mathrm{~mm}\) Hg to atmospheres, (d) \(1.323 \times 10^{5}\) Pa to atmospheres, (e) 2.50 atm to psi.

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gas state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.

Briefly explain the significance of the constants \(a\) and \(b\) in the van der Waals equation.

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