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Chapter 5: Problem 87

A gas evolved from the fermentation of glucose is found to effuse through a porous barrier in \(15.0 \mathrm{~min} .\) Under the same conditions of temperature and pressure, it takes an equal volume of \(\mathrm{N}_{2} 12.0 \mathrm{~min}\) to effuse through the same barrier. Calculate the molar mass of the gas and suggest what the gas might be.

Short Answer

Expert verified
The molar mass of the unknown gas is calculated to be approximately 32 g/mol, suggesting that the gas is likely to be Oxygen (O2).

Step by step solution

01

Use Graham's Law of effusion

Given that \( t_{1} = 15.0 min \) for the unknown gas and \( t_{2} = 12.0 min \) for Nitrogen gas, Graham's law is represented as \( \frac{t_{1}}{t_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \), where \(M_{1}\) is the molar mass of the first gas and \(M_{2}\) is the molar mass of the second gas. Substituting the times into the equation, we get \( \frac{15.0 min}{12.0 min} = \sqrt{\frac{M_{2}}{M_{1}}} \).
02

Calculate ratio of molar masses

Solve for \(\frac{M_{2}}{M_{1}}\) by squaring both sides of the equation. \(\left(\frac{15.0}{12.0}\right)^{2} = \frac{M_{2}}{M_{1}}\). Calculate \(\left(\frac{15.0}{12.0}\right)^{2}\) to find the ratio of the molar masses.
03

Calculate the molar mass of the unknown gas

Since we know the molar mass of Nitrogen gas \( (M_{1} = 28.0 g/mol) \), we can use the ratio of the molar masses determined in Step 2 to find the molar mass of the unknown gas. That is, \( M_{2} = (\frac{M_{2}}{M_{1}}) \cdot M_{1} \). Calculate the product to find the molar mass \( M_{2} \) of the unknown gas.
04

Identify the unknown gas

Based on the calculated molar mass, consult a list of gases and their molar masses to suggest what the unknown gas could be.

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Key Concepts

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

Effusion of Gases
Effusion is a process where gas molecules escape through a tiny hole into a vacuum. It's much like when air slowly leaks from a balloon with a tiny puncture. The rate at which this happens can tell us something about the nature of the gas itself.

In scientific terms, Graham's Law of effusion states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. What does this mean in simple terms? It's like comparing runners with different body weights; the lighter one will usually be faster. Similarly, lighter gases effuse more quickly than heavier ones.
Molar Mass Calculation
Molar mass is basically how much one mole of a substance weighs, and it's measured in grams per mole (g/mol). Think of it as a way to 'weigh' atoms and molecules since they're too small to put on a regular scale.

When you calculate molar mass, you're adding up the weight of all the atoms in a molecule. For example, water (H2O) has a molar mass of roughly 18.02 g/mol because it has two hydrogen atoms and one oxygen atom. To find the molar mass of a gas using effusion data, we use Graham's Law, as seen in the exercise solution.
Gas Laws
The behavior of gases is often predictable and can be described by gas laws — sets of mathematical equations. These include laws like Boyle's Law, which tells us about the relationship between pressure and volume, and Charles's Law, which talks about volume and temperature.

As for our problem, we've looked at how the gas behaves when it's effusing. By using the gas laws alongside Graham's Law of effusion, we can gain insights into the characteristics of a gas, such as its molar mass, by observing how it moves through a barrier compared to another known gas.
Fermentation of Glucose
Fermentation is a fascinating biological process where microorganisms like yeast transform carbohydrates like glucose into energy, often producing gases as a byproduct. The most commonly known products of glucose fermentation are ethanol and carbon dioxide (CO2).

In our exercise, the gas produced through the fermentation of glucose could be CO2, a common byproduct. This assumption can be checked by calculating the molar mass of the effused gas and comparing it to known values for CO2.

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Most popular questions from this chapter

Commercially, compressed oxygen is sold in metal cylinders. If a 120-L cylinder is filled with oxygen to a pressure of 132 atm at \(22^{\circ} \mathrm{C},\) what is the mass (in grams) of \(\mathrm{O}_{2}\) present? How many liters of \(\mathrm{O}_{2}\) gas at 1.00 atm and \(22^{\circ} \mathrm{C}\) could the cylinder produce? (Assume ideal behavior.)

Under the same conditions of temperature and pressure, why does \(1 \mathrm{~L}\) of moist air weigh less than \(1 \mathrm{~L}\) of dry air? In weather forecasts, an oncoming lowpressure front usually means imminent rainfall. Explain.

A sample of nitrogen gas kept in a container of volume \(2.3 \mathrm{~L}\) and at a temperature of \(32^{\circ} \mathrm{C}\) exerts a pressure of 4.7 atm. Calculate the number of moles of gas present.

A piece of sodium metal reacts completely with water as follows: $$ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) $$ The hydrogen gas generated is collected over water at \(25.0^{\circ} \mathrm{C}\). The volume of the gas is \(246 \mathrm{~mL}\) measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at \(25^{\circ} \mathrm{C}=0.0313\) atm. \()\)

Some commercial drain cleaners contain a mixture of sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: $$ \begin{array}{r} 2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ 2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g) \end{array} $$ The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of \(\mathrm{H}_{2}\) formed at \(23^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) if \(3.12 \mathrm{~g}\) of \(\mathrm{Al}\) are treated with an excess of \(\mathrm{NaOH}\).
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