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Equal masses of oxygen, hydrogen and methane are taken in identical conditions. What is the ratio of the volumes of the gases under identical conditions? (a) \(16: 1: 8\) (b) \(1: 16: 2\) (c) \(1: 16: 8\) (d) \(2: 16: 1\)

Short Answer

Expert verified
The ratio of the volumes of oxygen, hydrogen, and methane under identical conditions is (b) 1:16:2.

Step by step solution

01

Understanding Avogadro's Law

Avogadro's Law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. This implies that under identical conditions, the volume of a gas is directly proportional to the number of moles of the gas.
02

Determining Molar Masses

To find the ratio of the volumes, it is important to note the molar mass of each gas. Oxygen has a molar mass of 32 g/mol (since it is diatomic - O2), hydrogen has a molar mass of 2 g/mol (also diatomic - H2), and methane (CH4) has a molar mass of 16 g/mol.
03

Calculating Moles of Each Gas

Since equal masses of each gas are used, the number of moles of each gas can be determined by dividing the given mass by the molar mass. Let the common mass of each gas be 'm' g. Therefore, the number of moles of oxygen is m/32, the number of moles of hydrogen is m/2, and the number of moles of methane is m/16.
04

Applying Avogadro's Law to Find Volume Ratios

According to Avogadro's Law, volumes are proportional to moles. Hence, the volume ratio of oxygen : hydrogen : methane will be in the same ratio as their moles. Therefore, the volume ratio is (m/32) : (m/2) : (m/16).
05

Simplifying the Ratio

To find the simplest whole number ratio, divide each term by the smallest number of moles, which is m/32. Simplifying gives the ratio 1:16:2. Hence, the ratio of the volumes of oxygen, hydrogen, and methane under identical conditions is 1:16:2.

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Key Concepts

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

Molar Mass
When discussing gases and how they relate to chemical reactions and physical processes, molar mass serves as a fundamental concept. Molar mass is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The significance of molar mass lies in its ability to relate mass to moles, an essential conversion when exploring the behavior of gases with Avogadro's Law.

For instance, in our textbook exercise, different gases - oxygen, hydrogen, and methane, have different molar masses: 32 g/mol for oxygen (O2), 2 g/mol for hydrogen (H2), and 16 g/mol for methane (CH4). These values are crucial because they allow us to calculate the number of moles in a given mass of each gas. As the molar mass increases, fewer moles of the substance are present in the same mass. This concept is pivotal to understanding the relationships between the volume of gases and the amount of substance when discussing gas laws.
Volume Ratio of Gases
The volume ratio of gases is a direct outcome of Avogadro's Law, which states that equal volumes of gases, under the same conditions of temperature and pressure, contain equal numbers of molecules. Therefore, the volume of a gas is proportional, not to the mass of the gas, but to the number of moles of the gas. This is a key concept when comparing different gases under the same conditions.

Using our exercise as an example, once the molar masses are known, and equal masses of these gases are considered, we can determine the volume ratio by first calculating the moles of each gas. From there, it's a straightforward step to apply the proportional relationship between moles and volume to find these ratios. This approach allows us to compare different gases fairly, irrespective of their unique chemical compositions or properties.
Moles of Gas
Moles of gas is a unit of measurement that represents the amount of a substance. One mole contains exactly 6.022 x 1023 particles, known as Avogadro's number. In the context of gas laws, the number of moles indicates the quantity of gas molecules present, regardless of the type of gas. This quantity directly corresponds to the volume a gas occupies under standard conditions as described by Avogadro's Law.

In our exercise, the importance of understanding moles of gas is underscored when equal masses of different gases are considered. By dividing the mass of each gas by its respective molar mass, we obtain the number of moles. This step is pivotal for applying Avogadro's Law to determine the volume ratio. The calculations provide a clear perspective on how different gases, while having the same mass, occupy volumes proportional to their number of moles in identical conditions of pressure and temperature.

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Most popular questions from this chapter

Some bottles of colourless liquids were being labelled when the technicians accidentally mixed them up and lost track of their contents. A \(15.0 \mathrm{ml}\) sample withdrawn from one bottle weighed \(22.3 \mathrm{~g} .\) The technicians knew that the liquid was either acetone, benzene, chloroform on carbon tetrachloride(which have densities of \(0.792 \mathrm{~g} / \mathrm{cm}^{3}, 0.899 \mathrm{~g} / \mathrm{cm}^{3}, 1.489 \mathrm{~g} / \mathrm{cm}^{3}\), and \(1.595 \mathrm{~g} / \mathrm{cm}^{3}\), respectively). What was the identity of the liquid? (a) Carbon tetrachloride (b) Acetone (c) Chloroform (d) Benzene

Total number of electrons present in \(4.4 \mathrm{~g}\) oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right)\) is (a) \(0.05 N_{\mathrm{A}}\) (b) \(2.3 N_{\mathrm{A}}\) (c) \(2.2 N_{\mathrm{A}}\) (d) \(2.1 N_{\mathrm{A}}\)

How much \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\), in \(\mathrm{mg}\), must be present in \(50 \mathrm{ml}\) of a solution with \(2.35\) ppm of Ca? (a) \(0.1175\) (b) \(770.8\) (c) \(4.7\) (d) \(0.48\)

To determine soluble (free) \(\mathrm{SiO}_{2}\) in a rock. an alkaline extraction was carried out, as a result of which there was found \(1.52 \%\) of \(\mathrm{SiO}_{2}\) in the extract and also \(1.02 \%\) of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Considering that, apart from the free \(\mathrm{SiO}_{2}\), the extract also contained the \(\mathrm{SiO}_{2}\) that had passed into it from Kaolin \(\left(2 \mathrm{SiO}_{2} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}\right)\), the percentage of free \(\mathrm{SiO}_{2}\) in the rock being analysed is \((\mathrm{Si}=28\), \(\mathrm{Al}=27\) ) (a) \(1.20\) (b) \(0.32\) (c) \(0.50\) (d) \(1.52\)

A compound having the empirical formula, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\), has a molecular weight of \(170 \pm 5\). The molecular formula of the compound is (a) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{3}\)

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