Question

An unknown gas ' \(\mathrm{X}\) ' has rate of diffusion measured to be \(0.88\) times that of \(\mathrm{PH}_{3}\) at the same conditions of temperature and pressure. The gas may be (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) \(\mathrm{CO}\) (c) \(\mathrm{NO}_{2}\) (d) \(\mathrm{N}_{2} \mathrm{O}\)

Short answer

The unknown gas \(\mathrm{X}\) is \(\mathrm{N}_{2} \(\mathrm{O}\)\).

Step-by-step solution

Apply Graham's Law of Effusion

According to Graham's Law of Effusion, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. The formula can be stated as Rate1 / Rate2 = sqrt(Molar Mass2 / Molar Mass1).

Set up the ratio using the given data

The problem states that the rate of gas X is 0.88 times the rate of \(\mathrm{PH}_{3}\). Therefore, we set up the ratio as 0.88 = sqrt(Molar Mass of \(\mathrm{PH}_{3}\) / Molar Mass of X).

Calculate the molar mass of \(\mathrm{PH}_{3}\)

To find the molar mass of \(\mathrm{PH}_{3}\), add the atomic masses of 1 phosphorus atom and 3 hydrogen atoms. Phosphorus has an approximate atomic mass of 31 g/mol, and hydrogen has an approximate atomic mass of 1 g/mol. The molar mass of \(\mathrm{PH}_{3}\) is 31 + 3(1) = 34 g/mol.

Determine the Molar Mass of gas X

Using the equation from Step 2, solve for the Molar Mass of X. 0.88^2 = 34 / Molar Mass of X. Molar Mass of X = 34 / 0.88^2.

Calculate the square and solve for Molar Mass of X

0.88^2 equals approximately 0.7744. Now, divide 34 g/mol by 0.7744 to get the Molar Mass of X, which is approximately 43.9 g/mol.

Identify the correct gas

Compare the calculated molar mass with the molar masses of the given options: (a) \(\mathrm{C}_{2} \(\mathrm{H}_{6}\)\) has a molar mass of 30 g/mol (2 carbon at 12 g/mol each + 6 hydrogen at 1 g/mol each). (b) \(\mathrm{CO}\) has a molar mass of 28 g/mol (1 carbon at 12 g/mol + 1 oxygen at 16 g/mol). (c) \(\mathrm{NO}_{2}\) has a molar mass of 46 g/mol (1 nitrogen at 14 g/mol + 2 oxygen at 16 g/mol each). (d) \(\mathrm{N}_{2} \(\mathrm{O}\)\) has a molar mass of 44 g/mol (2 nitrogen at 14 g/mol each + 1 oxygen at 16 g/mol). The closest molar mass to our calculated value is that of \(\mathrm{N}_{2} \(\mathrm{O}\)\), making option (d) the correct choice.

Key concepts

Rate of Diffusion

The rate of diffusion of a gas refers to how quickly it spreads out and mixes with another gas. It is an important concept that explains how gases move from areas of higher concentration to areas of lower concentration.

Graham's Law of Effusion states that the rate is inversely proportional to the square root of its molar mass, which can be expressed mathematically as the relationship: \( \text{Rate}_{1} / \text{Rate}_{2} = \sqrt{\frac{\text{Molar Mass}_{2}}{\text{Molar Mass}_{1}}} \). This means that lighter gases diffuse more rapidly than heavier ones. For instance, if we compare helium to oxygen, helium will diffuse faster due to its lower molar mass.

In our textbook problem, the unknown gas diffuses at a rate that is 0.88 times the rate of \(\mathrm{PH}_{3}\). Applying Graham’s Law will help us to find the molar mass of the unknown gas by comparing it to the known rate of \(\mathrm{PH}_{3}\).

Molar Mass Calculation

Understanding how to calculate molar mass is vital when working with gases, as it tells us how much one mole of a given substance weighs. The molar mass is the sum of the atomic masses of all the atoms in a molecule.

For instance, the molar mass of \(\mathrm{PH}_{3}\) is calculated by adding the atomic mass of phosphorus (\(31 \text{ g/mol}\)) and three times the atomic mass of hydrogen (\(3 \times 1 \text{ g/mol}\)). This gives us a molar mass of \(34 \text{ g/mol}\) for \(\mathrm{PH}_{3}\).

Using the molar masses, we can then solve Graham’s Law equations to find the molar mass of unknown gases. With the ratio of the rates of diffusion, we determine an unknown gas’s molar mass which is crucial for identification.

Gas Identification

Identifying a gas can be done by using its physical properties, one of which is the molar mass. In our exercise, the problem provides a list of potential gases, and with the calculated molar mass of the unknown gas, we can identify it by comparing it to the known molar masses of the given options.

For accurate gas identification, it is essential to have a value that closely matches one of the possible choices. In our case, the unknown gas has an approximate molar mass of \(43.9 \text{ g/mol}\) which is closest to the molar mass of \(\mathrm{N}_{2}\mathrm{O}\), allowing us to conclude it matches our unknown gas.

Comparing Molar Masses

To compare molar masses, it is essential first to understand the concept of molar mass and to know the atomic masses of elements, which are typically given in units of grams per mole (g/mol).

Using these atomic masses, we develop an understanding of relative weights of different molecules. In our exercise, the calculated molar mass of the unknown gas (approximately \(43.9 \text{ g/mol}\)) is compared with the molar masses of the options given in the problem.

It's crucial to remember that slight differences in molar masses can result in identification errors. Therefore, we look for the option with a molar mass that is the closest to our calculated value to determine the identity of the unknown gas. With \(\mathrm{N}_{2}\mathrm{O}\)'s molar mass at 44 g/mol, it's a close match and can be confidently identified as the unknown gas.