Question

The rate of diffusion of two gases \(X\) and \(\mathrm{Y}\) is in the ratio \(1: 5\) and that of \(\mathrm{Y}\) and \(\mathrm{Z}\) in the ratio of \(1: 6 .\) The ratio of the rate of diffusion of \(\mathrm{Z}\) with respect to \(\mathrm{X}\) is (a) \(30 / 1\) (b) \(1 / 30\) (c) \(5 / 6\) (d) \(6 / 5\)

Short answer

The ratio of the rate of diffusion of Z with respect to X is 30:1.

Step-by-step solution

Understanding the problem

We are given the ratios of rates of diffusion of two pairs of gases. According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The rate ratio for gases X and Y is 1:5 and for Y and Z is 1:6. We need to find the ratio of the rate of diffusion of Z to X.

Express the given ratios using variables

Let the rate of diffusion of gas X be denoted by Rx, that of Y by Ry, and that of Z by Rz. We can express the ratios using these variables: Rx : Ry = 1:5 and Ry : Rz = 1:6. We need to find Rz : Rx.

Calculating the ratio of Z to X

To find the ratio Rz : Rx, we first find the value of Ry in terms of Rx using the first ratio, which gives us Ry = 5Rx. Then, using the second ratio, we find Rz in terms of Ry, which gives us Rz = 6Ry. Substituting the value of Ry into this expression, we get Rz = 6(5Rx) = 30Rx. Thus, the ratio Rz : Rx is 30Rx : Rx, which simplifies to 30:1.

Key concepts

Graham's Law of Diffusion

Understanding the rate at which gases diffuse, or spread out, is critical for many scientific and industrial processes. Graham's Law of Diffusion provides an insight into this concept, stating that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. In simpler terms, a lighter gas will diffuse more rapidly than a heavier one. This fundamental principle can be expressed mathematically as:

\( \text{rate of diffusion} \propto \frac{1}{{\sqrt{\text{molar mass}}}} \).

It's essential to understand that the 'rate of diffusion' in this context refers to the speed at which a gas spreads through another or within a given space. This law can be applied to explain and predict how different gases will behave under the same conditions. For example, helium will diffuse faster than oxygen because it has a lower molar mass. For students grappling with problems involving gas diffusion, connecting the rate directly to the molar mass through this equation is highly beneficial for accurate and simplified problem solving.

Molar Mass

The molar mass of a substance is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The unit for molar mass is grams per mole (g/mol). It can be calculated by summing the masses of the individual atoms within a molecule, based on the atomic mass of each element as found on the periodic table. For instance, the molar mass of water (H2O) is calculated as follows:

\( 2(\text{atomic mass of H}) + \text{atomic mass of O} \)

Together, this sums to approximately 18 g/mol. Molar mass is crucial when studying the properties of gases because it directly relates to how gases diffuse, as indicated by Graham's Law. To put it into perspective, comparing the molar masses of different gases allows us to predict how quickly or slowly they will diffuse. If a student knows the molar masses of two gases and their rate of diffusion through Graham's Law, they can solve for various unknowns in gas behavior problems.

Gas Diffusion Ratios

In the realm of gases, diffusion ratios are a way of expressing how the rates of diffusion of two gases compare to each other. These ratios are grounded in the principles of Graham's Law and depend on the molar masses of the gases involved. The rate ratio of two gases can be written as:

\( \frac{\text{Rate of gas A}}{\text{Rate of gas B}} = \sqrt{\frac{\text{Molar mass of B}}{\text{Molar mass of A}}} \).

This equation tells us that if you have two gases, A and B, the lighter gas (with the lower molar mass) will have the greater diffusion rate. Gas diffusion ratios can therefore help predict how gases will interact and spread in different settings. For example, in the exercise, the given diffusion ratio of gas X to gas Y is 1:5, which implies that gas Y diffuses five times faster than gas X. Similarly, the 1:6 ratio between gas Y and gas Z indicates that gas Z diffuses six times faster than gas Y. By understanding and applying these ratios, students can solve complex gas diffusion problems systematically, making challenging concepts more approachable.