Equal volumes of oxygen gas and a second gas weigh \(1.00\) and \(2.375\) grams respectively under the same experimental conditions. Which of the following is the unknown gas? (a) \(\mathrm{NO}\) (b) \(\mathrm{SO}_{2}\) (c) \(\mathrm{CS}_{2}\) (d) \(\mathrm{CO}\)

Short Answer

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(c) CS2

Step by step solution

01

Determine the molar mass of oxygen gas

Oxygen gas (O2) has a molar mass of 16.00 g/mol for each oxygen atom. Since oxygen gas is diatomic (O2), its molar mass is twice that of an oxygen atom. Therefore, the molar mass of O2 is 16.00 g/mol x 2 = 32.00 g/mol.
02

Calculate the molar mass of the unknown gas

Given equal volumes and conditions of two gases, according to Avogadro's law, they contain the same number of moles. If 1.00 gram of O2 is equivalent to 1.00 / 32.00 moles of O2, the unknown gas weighing 2.375 grams will also have that many moles. To find the molar mass of the unknown gas, divide its weight by the number of moles: Molar mass of the unknown gas = 2.375 grams / (1.00 / 32.00 moles) = 2.375 grams x (32.00 / 1.00) = 76.00 g/mol.
03

Identify the unknown gas by its molar mass

Compare the calculated molar mass of the unknown gas to the molar masses of the options provided: (a) NO (30.01 g/mol), (b) SO2 (64.07 g/mol), (c) CS2 (76.14 g/mol), (d) CO (28.01 g/mol). The molar mass of the unknown gas (76.00 g/mol) is closest to the molar mass of CS2 (76.14 g/mol), which makes (c) CS2 the correct answer.

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Key Concepts

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

Understanding Avogadro's Law
Avogadro's Law is a fundamental principle in physical chemistry which states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. This concept is vital when comparing different gases as it allows scientists and students to predict how gases will behave under similar conditions.

When we use Avogadro's Law for calculations, what we are essentially saying is that the volume of a gas is directly proportional to the number of moles (a measure of quantity in chemistry) present, as long as the temperature and pressure remain constant. To put it into a simple equation form, we could write it as \( V \propto n \), where \( V \) represents the volume and \( n \) represents the number of moles.

Applying this to our textbook problem simplifies our task: since the volumes of oxygen and the unknown gas are equal and under the same conditions, they must contain the same number of moles. This insight is crucial in solving for the molar mass of the unknown gas, leading us to identify the gas itself.
Molar Mass Calculation Explained
Molar mass is the weight of one mole of a given substance and is expressed in grams per mole (g/mol). It is a bridge between the macroscopic world we can measure and the microscopic atomic world. The molar mass of an element can be found on the periodic table as it is equivalent to the atomic weight of the element.

To calculate the molar mass of a compound, we sum up the molar masses of all the atoms that make up the compound. For diatomic oxygen (O2), this involves adding the molar masses of two oxygen atoms. Since the atomic weight of oxygen is approximately 16.00 g/mol, diatomic oxygen has a molar mass of \( 16.00 \, g/mol \times 2 = 32.00 \, g/mol \).

Using the molar mass, we can convert between mass and moles, allowing us to perform a diverse array of chemical calculations. This is displayed in the solution where, after determining the moles of oxygen based on its mass, we used Avogadro's Law and the mass of the unknown gas to find its molar mass.
Comparison of Gases in Chemical Equations

Gases on a Balanced Scale

Comparing gases is a common task in physical chemistry, often required for reactions involving gaseous reactants or products. When comparing gases, we need to look at their molar masses and understand how they relate to their behaviors under the same conditions of temperature and pressure.

Role of Molar Mass in Identifying Gases

In situations like our exercise, once we know the molar mass of an unknown gas, we can use it to determine the gas's identity by comparing it with known values. It's like being a detective, where the molar mass is a clue that leads us to the name of the gas.

In our exercise, with the molar mass of the mystery gas at hand, we compared it to the molar masses of potential candidates. This comparison, much like a lineup, enabled us to pinpoint the correct gas with a molar mass that matched our calculated value.

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Most popular questions from this chapter

The density of gas \(A\) is twice that to \(B\) at the same temperature the molecular weight of gas \(B\) is twice that of \(A\). The ratio of pressure of gas \(A\) and \(B\) will be : (a) \(1: 6\) (b) \(1: 1\) (c) \(4: 1\) (d) \(1: 4\)

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At low pressure, the van der Waals equation become : (a) \(P V_{m}=R T\) (b) \(P\left(V_{m}-b\right)=R T\) (c) \(\left(P+\frac{a}{V_{m}^{2}}\right) V_{m}=R T\) (d) \(P=\frac{R T}{V_{m}}+\frac{a}{V_{m}^{2}}\)

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The density of liquid gallium at \(30^{\circ} \mathrm{C}\) is \(6.095 \mathrm{~g} / \mathrm{mL}\). Because of its wide liquid range ( 30 to \(2400^{\circ} \mathrm{C}\) ), gallium could be used as a barometer fluid at high temperature. What height (in \(\mathrm{cm}\) ) of gallium will be supported on a day when the mercury barometer reads 740 torr? (The density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\).). (a) 322 (b) 285 (c) 165 (d) 210

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